Nonamenable simple $C^*$-algebras with tracial approximation

Abstract

We construct two types of unital separable simple $C^*$ -algebras: $A_z^{C_1}$ and $A_z^{C_2}$ , one exact but not amenable, the other nonexact. Both have the same Elliott invariant as the Jiang–Su algebra – namely, $A_z^{C_i}$ has a unique tracial state,

$$ \begin{align*} \left(K_0\left(A_z^{C_i}\right), K_0\left(A_z^{C_i}\right)_+, \left[1_{A_z^{C_i}} \right]\right)=(\mathbb{Z}, \mathbb{Z}_+,1), \end{align*} $$

and $K_{1}\left (A_z^{C_i}\right )=\{0\}$ ( $i=1,2$ ). We show that $A_z^{C_i}$ ( $i=1,2$ ) is essentially tracially in the class of separable ${\mathscr Z}$ -stable $C^*$ -algebras of nuclear dimension $1$ . $A_z^{C_i}$ has stable rank one, strict comparison for positive elements and no $2$ -quasitrace other than the unique tracial state. We also produce models of unital separable simple nonexact (exact but not nuclear) $C^*$ -algebras which are essentially tracially in the class of simple separable nuclear ${\mathscr Z}$ -stable $C^*$ -algebras, and the models exhaust all possible weakly unperforated Elliott invariants. We also discuss some basic properties of essential tracial approximation.

1. Introduction

Simple unital projectionless amenable $C^*$ -algebras were first constructed by Blackadar [2]. The $C^*$ -algebra A constructed by Blackadar has the property that $K_0(A)=\mathbb {Z}$ with the usual order but with nontrivial $K_1(A)$ . The Jiang–Su algebra ${\mathscr Z}$ given by Jiang and Su [27] is a unital infinite-dimensional separable amenable simple $C^*$ -algebra with Elliott invariant exactly the same as that of the complex field $\mathbb {C}$ , Let A be any $\sigma $ -unital $C^*$ -algebra. Then $K_i(A)=K_i(A\otimes {\mathscr Z})$ ( $i=0,1$ ) as abelian groups and $T(A)\cong T(A\otimes {\mathscr Z})$ . If A is a separable simple $C^*$ -algebra, then $A\otimes {\mathscr Z}$ has nice regularity properties. For example, $A\otimes {\mathscr Z}$ is either purely infinite or stably finite [42]. In fact, if $A\otimes {\mathscr Z}$ is not purely infinite, then it has stable rank one when A is not stably projectionless [42], or it almost has stable rank one when it is stably projectionless [38]. Also, $A\otimes {\mathscr Z}$ has weakly unperforated $K_0$ -group [23]. Another important regularity property is that $A\otimes {\mathscr Z}$ has strict comparison [42] (see also Definition 2.6). If A has weakly unperforated $K_0(A)$ , then A and $A\otimes {\mathscr Z}$ have the same Elliott invariant. In other words, A and $A\otimes {\mathscr Z}$ are not distinguishable from the Elliott invariant.

The Jiang–Su algebra ${\mathscr Z}$ is an inductive limit of $1$ -dimensional noncommutative CW complexes. In fact, ${\mathscr Z}$ is the unique infinite-dimensional separable simple $C^*$ -algebra with finite nuclear dimension in the UCT class which has the same Elliott invariant as that of the complex field $\mathbb {C}$ (see [16, Corollary 4.12]). These properties give ${\mathscr Z}$ a prominent role in the study of structure of $C^*$ -algebras, in particular in the study of classification of amenable simple $C^*$ -algebras.

Attempts to construct a nonnuclear Jiang–Su-type $C^*$ -algebra have been on the horizon for over a decade. In particular, after Dădărlat’s construction of nonamenable models for non–type I separable unital AF-algebras [13], this should be possible. The construction in [13] generalised some earlier constructions of simple $C^*$ -algebras of real rank zero such as that of Goodearl [25]. Jiang and Su’s construction has a quite different feature. To avoid producing any nontrivial projections, unlike Dădărlat’s construction, Jiang and Su did not use any finite-dimensional representations as a direct summand of connecting maps in the inductive systems. The construction used prime-dimension drop algebras, and connecting maps were highly inventive so that the traces eventually collapse to one. In fact, Rørdam and Winter took another approach [43] using a $C^*$ -subalgebra of $C\!\left([0,1], M_{\mathfrak {p}}\otimes M_{\mathfrak {q}}\right )$ , where $\mathfrak {p}$ and $\mathfrak {q}$ are relatively prime supernatural numbers. One possible attempt to construct a nonamenable Jiang–Su-type $C^*$ -algebra would use $C\!\left([0,1], B_{\mathfrak {p}}\otimes B_{{\mathfrak {q}}}\right )$ , where $B_{\mathfrak {p}}$ and $B_{\mathfrak {q}}$ are, respectively, nonamenable models for $M_{\mathfrak {p}}$ and $M_{\mathfrak {q}}$ constructed in [13]. However, one usually would avoid computation of the K-theory of tensor products of nonexact simple $C^*$ -algebras such as $B_{\mathfrak {p}}$ and $B_{\mathfrak {q}}$ . Moreover, Rørdam and Winter’s construction depends on knowing the existence of the Jiang–Su algebra ${\mathscr Z}$ . On the other hand, if one considers nonexact interval ‘dimension drop algebras’, besides controlling K-theory one has additional issues such as the fact that each fibre of the ‘dimension drop algebra’ is not simple (unlike the usual dimension drop algebras, whose fibres are simple matrix algebras).

We will present some nonexact (or exact but nonnuclear) unital separable simple $C^*$ -algebras $A_z^C$ which have the property that their Elliott invariants are the same as that of the Jiang–Su algebra ${\mathscr Z}$ – namely, $\left (K_0\left (A_z^C\right ), K_0\left (A_z^C\right )_+, \left [1_{A_z^C}\right ]\right )=(\mathbb {Z}, \mathbb {Z}_+,1)$ , $K_1\left (A_z^C\right ) =\{0\}$ and $A_z^C$ has a unique tracial state. Moreover, $A_z^C$ has stable rank one and has strict comparison for positive elements. $A_z^C$ has no (nonzero) $2$ -quasitrace other than the unique tracial state. Even though $A_z^C$ may not be exact, it is essentially tracially approximated by ${\mathscr Z}$ . In particular, it is essentially tracially approximated by unital simple $C^*$ -algebras with nuclear dimension $1$ .

In this paper, we will also study the tracial approximation. We will make it precise what we mean by saying that $A_z^C$ is essentially tracially approximated by ${\mathscr Z}$ (Definition 3.1, Lemma 8.1). We expect that regularity properties such as stable rank one, strict comparison for positive elements and almost unperforated Cuntz semigroups, as well as approximate divisibility, are preserved by tracial approximation. In fact, we show that if a unital separable simple $C^*$ -algebra A is essentially tracially in ${\cal C}_{\mathscr Z}$ , the class of ${\mathscr Z}$ -stable $C^*$ -algebras, then – as far as the usual regularity properties are concerned – A behaves just like $C^*$ -algebras in ${\cal C}_{\mathscr Z}$ . More precisely, we show that if A is simple and essentially tracially in ${\cal C}_{\mathscr Z}$ , then A is tracially approximately divisible. If A is not purely infinite, then it has stable rank one (or almost has stable rank one, if A is not unital) and has strict comparison, and its Cuntz semigroup is almost unperforated. If A is essentially tracially in the class of exact $C^*$ -algebras, then every $2$ -quasitrace of $\overline {aAa}$ , for any a in the Pedersen ideal of A, is in fact a trace.

Using $A_z^C$ , we present a large class of nonexact (or exact but nonnuclear) unital separable simple $C^*$ -algebras which exhaust all possible weakly unperforated Elliott invariants. Moreover, every $C^*$ -algebra in the class is essentially tracially in the class of unital separable simple $C^*$ -algebras which are ${\mathscr Z}$ -stable, and has nuclear dimension at most $1$ .

The paper is organised as follows: Section 2 serves as preliminaries, where some frequently used notations and definitions are listed. Section 3 introduces the notion of essential tracial approximation for simple $C^*$ -algebras. In Section 4 we present some basic properties of essential tracial approximation. For example, we show that if A is a simple $C^*$ -algebra and is essentially tracially approximated by $C^*$ -algebras whose Cuntz semigroups are almost unperforated, then the Cuntz semigroup of A is almost unperforated (Theorem 4.3). In particular, A has strict comparison for positive elements. In Section 5 we study the separable simple $C^*$ -algebras which are essentially tracially approximated by ${\mathscr Z}$ -stable $C^*$ -algebras. We show that such $C^*$ -algebras are either purely infinite or almost have stable rank one (or do have stable rank one, if the $C^*$ -algebras are unital). These simple $C^*$ -algebras are tracially approximately divisible and have strict comparison for positive elements. In Section 6 we begin the construction of $A_z^C$ . In Section 7 we show that the construction in Section 6 can be made simple, and the Elliott invariant of $A_z^C$ is precisely the same as that of a complex field, just as with the Jiang–Su algebra ${\mathscr Z}$ . In Section 8 we show that $A_z^C$ has all expected regularity properties. Moreover, $A_z^C$ is essentially tracially approximated by ${\mathscr Z}$ . Using $A_z^C$ , we also produce, for each weakly unperforated Elliott invariant, a unital separable simple nonexact (or exact but nonnuclear) $C^*$ -algebra B which has the said Elliott invariant, has stable rank one, is essentially tracially approximated by $C^*$ -algebras with nuclear dimension at most $1$ , has almost unperforated Cuntz semigroup, has strict comparison for positive elements and has no $2$ -quasitraces which are not traces.

2. Preliminaries

In this paper, the set of all positive integers is denoted by $\mathbb {N}$ . If A is unital, $U(A)$ is the unitary group of A. A linear map is said to be c.p.c., if it is a completely positive contraction.

Notation 2.1. Let A be a $C^*$ -algebra and ${\cal F}\subset A$ be a subset. Let $\epsilon>0$ . Set $a,b\in A$ and write $a\approx _{\epsilon }b$ if $\lVert a-b\rVert < \epsilon $ . We write $a\in _\varepsilon {\cal F}$ if there is $x\in {\cal F}$ such that $a\approx _\varepsilon x$ .

Notation 2.2. Let A be a $C^*$ -algebra and let $S\subset A$ be a subset of A. Denote by $\mathrm {Her}_A(S)$ (or just $\mathrm {Her}(S)$ , when A is clear) the hereditary $C^*$ -subalgebra of A generated by S. Denote by $A^{\textbf{1}}$ the closed unit ball of A, by $A_+$ the set of all positive elements in A, by $A_+^{\textbf{1}}:=A_+\cap A^{\textbf{1}}$ and by $A_{\mathrm {sa}}$ the set of all self-adjoint elements in A. Denote by $\widetilde A$ (or $A^{\sim }$ ) the minimal unitisation of A. When A is unital, denote by $GL(A)$ the set of invertible elements of A and by $U(A)$ the unitary group of A.

Notation 2.3. Let $\epsilon>0$ . Define a continuous function $f_{\epsilon }: {{[0,+\infty )}} \rightarrow [0,1]$ by

$$ \begin{align*} f_{\epsilon}(t) := \begin{cases} 0, &t\in {{[0,\epsilon/2]}},\\ 1, &t\in [\epsilon,\infty),\\ \mathrm{linear}, &{t\in[\epsilon/2, \epsilon].} \end{cases} \end{align*} $$

Definition 2.4. Let A be a $C^*$ -algebra and set $M_{\infty }(A)_+:=\bigcup _{n\in \mathbb {N}}M_n(A)_+$ . For $x\in M_n(A)$ , we identify x with $\mathrm {diag}(x,0)\in M_{n+m}(A)$ for all $m\in \mathbb {N}$ . Set $a\in M_n(A)_+$ and $b\in M_m(A)_+$ . We may write $a\oplus b:=\mathrm {diag}(a,b)\in M_{n+m}(A)_+$ . If $a, b\in M_n(A)$ , we write $a \lesssim b$ if there are $x_i\in M_n(A)$ such that $\lim _{i\rightarrow \infty }\left \lVert a-x_i^*bx_i\right \rVert =0$ . We write $a \sim b$ if $a \lesssim b$ and $b \lesssim a$ hold. The Cuntz relation $\sim $ is an equivalence relation. Set $W(A):=M_{\infty }(A)_+/\sim $ . Let $\langle a\rangle $ denote the equivalence class of a. We write $\langle a\rangle \leq \langle b\rangle $ if $a \lesssim b$ . $(W(A),\leq )$ is a partially ordered abelian semigroup. Let $\mathrm {Cu}(A)=W(A\otimes {\cal K})$ . $W(A)$ (resp., $\mathrm {Cu}(A)$ ) is called almost unperforated if, for any $\langle a \rangle , \langle b\rangle \in W(A)$ (resp., $\mathrm {Cu}(A)$ ) and any $k\in \mathbb {N}$ , when $(k+1)\langle a\rangle \leq k\langle b\rangle $ , we have $\langle a \rangle \leq \langle b\rangle $ (see [40]).

Let $B\subset A$ be a hereditary $C^*$ -subalgebra, and set $a,b\in B_+$ . It is clear that $a\lesssim _B b$ implies $a\lesssim _A b$ . Conversely, if $a\lesssim _A b$ , then, for any $\varepsilon>0$ , there exists $x\in A$ such that $\left \lVert a-x^*bx\right \rVert <\varepsilon /4$ . Choose $e\in B_+^{\textbf{1}}$ such that $\lVert a-eae\rVert <\varepsilon /4$ . Then $\left \lVert a-ex^*b^{1/4} b^{1/2} b^{1/4} xe\right \rVert <\varepsilon /2$ . It follows that $a\lesssim _B b^{1/2}\sim _B b$ . In other words, $a\lesssim _A b \Leftrightarrow a\lesssim _B b$ .

Remark 2.5. It is known to some experts that the condition that $W(A)$ be almost unperforated is equivalent to the condition that $\mathrm {Cu}(A)$ be almost unperforated. To see this briefly, let us assume that $W(A)$ is almost unperforated and set $a, b\in (A\otimes {\cal K})_+$ such that $(k+1)\langle a\rangle \le k\langle b\rangle $ . Let $\left \{e_{i,j}\right \}$ be the system of matrix units for ${\cal K}$ and $E_n=\sum _{i=1}^n1_{\tilde A}\otimes e_{i,i}$ , and let $\varepsilon>0$ . Note that $E_naE_n\in M_n(A)_+$ for all $n\in \mathbb {N}$ . Moreover, $a\approx _{\varepsilon /8} E_naE_n$ for some large $n\in \mathbb {N}$ . It follows from [40, Proposition 2.2] that $(a-\varepsilon )_+\lesssim (E_naE_n-\varepsilon /4)_+$ and $(E_naE_n-\varepsilon /4)_+\lesssim (a-\varepsilon /8)_+$ . By [40, Proposition 2.4], there exists $\delta>0$ such that $(k+1)\langle (a-\varepsilon /8)_+\rangle \le k\langle (b-\delta )_+\rangle $ . Repeating Rørdam’s results [40], one obtains that $\langle (b-\delta )_+\rangle \le {{\langle E_mbE_m\rangle }}$ for some even larger m ( $m\ge n$ ). Now one has $(k+1)\langle (E_naE_n-\varepsilon /4)_+\rangle \le k\langle E_mbE_m\rangle $ . By the last paragraph of Definition 2.4, this holds in $M_m(A)$ . Since $W(A)$ is almost unperforated, $(a-\varepsilon )_+\lesssim (E_naE_n-\varepsilon /4)_+\lesssim E_mbE_m$ . Then $(a-\varepsilon )_+\lesssim E_mbE_m\lesssim b$ . It follows that $a\lesssim b$ . Therefore $W(A)$ being almost unperforated implies that $\mathrm {Cu}(A)$ is almost unperforated.

To see the converse, just notice again that A is a hereditary $C^*$ -subalgebra of $A\otimes {\cal K}$ ; then $\langle a\rangle \leq \langle b\rangle $ in $\mathrm {Cu}(A)=W(A\otimes {\cal K})$ implies $\langle a\rangle \leq \langle b\rangle $ in $W(A)$ .

Definition 2.6. Denote by $QT(A)$ the set of $2$ -quasitraces of A with $\lVert \tau \rVert =1$ (see [4, II 1.1, II 2.3]) and by $T(A)$ the set of all tracial states on A. We will also use $T(A)$ as well as $QT(A)$ for the extensions on $M_k(A)$ for each k. In fact, $T(A)$ and $QT(A)$ may be extended to lower semicontinuous traces and lower semicontinuous quasitraces on $A\otimes {\cal K}$ (see before [17, Proposition 4.2] and [7, Remark 2.27(viii)]).

Let A be a $C^*$ -algebra. Denote by $\mathrm {Ped}(A)$ the Pedersen ideal of A (see [36, 5.6]). Suppose that A is a $\sigma $ -unital simple $C^*$ -algebra. Choose $b\in \mathrm {Ped}(A)_+$ with $\lVert b\rVert =1$ . Put $B:=\overline {bAb}=\mathrm {Her}(b)$ . Then by [8], $A\otimes {\cal K}\cong B\otimes {\cal K}$ . For each $\tau \in QT(B)$ , define a lower semicontinuous function $d_\tau : A\otimes {\cal K}_+ \rightarrow [0,+\infty ]$ , $x\mapsto \lim _{n\rightarrow \infty }\tau \left (f_{1/n}(x)\right )$ . The function $d_{\tau }$ is called the dimension function induced by $\tau $ .

We say A has strict comparison (for positive elements) if, for any $a, b\in A\otimes {\cal K}_+$ , the statement $d_\tau (a)<d_\tau (b)$ for all $\tau \in QT(B)$ implies that $a\lesssim b$ .

3. Tracial approximation

Definition 3.1. Let ${\cal P}$ be a class of $C^*$ -algebras that is closed under isomorphisms, and let A be a simple $C^*$ -algebra. We say A is essentially tracially in ${\cal P}$ (abbreviated as ‘e. tracially in ${\cal P}$ ’) if, for any finite subset ${\cal F}\subset A$ , any $\varepsilon>0$ and any $s\in A_+\setminus \{0\}$ , there exist an element $e\in A_+^{\textbf{1}}$ and a nonzero $C^*$ -subalgebra B of A which is in ${\cal P}$ such that the following hold:

1. (1) $\lVert ex-xe\rVert <\varepsilon \text { for all } {{x}} \in {\cal F}$ .

2. (2) $(1-e)x\in _{\varepsilon } B$ and $\lVert (1-e)x\rVert \ge \lVert x\rVert -\varepsilon $ for all $x\in {\cal F}$ .

3. (3) $e\lesssim s$ .

Proposition 3.2. Let ${\cal P}$ be a class of $C^*$ -algebras and let A be a simple $C^*$ -algebra. Then A is e. tracially in ${\cal P}$ if and only if the following hold: For any $\varepsilon>0$ , any finite subset ${\cal F}\subset A$ , any $a \in A_+\setminus \{0\}$ and any finite subset ${\cal G}\subset C_0([0,1])$ , there exist an element $e\in A_+^{\textbf{1}}$ and a nonzero $C^*$ -subalgebra B of A such that B in $\mathcal {P}$ , and the following hold:

1. (1) $\lVert ex-xe\rVert <\varepsilon $ for all $ x\in \cal F$ .

2. (2) $g(1-e)x\in _\varepsilon B$ for all $g\in {\cal G}$ and $\lVert (1-e)x\rVert \ge \lVert x\rVert -\varepsilon $ for all $x\in {\cal F}$ . and

3. (3) $e\lesssim a$ .

Proof. The ‘if’ part follows easily by taking ${\cal G}=\{\iota \}$ , where $\iota (t)=t$ for all $t\in [0,1]$ .

We now show the ‘only if’ part.

Suppose that A is e. tracially in ${\cal P}$ . Let $\varepsilon>0$ and let ${\cal F}\subset A$ be a finite subset, and without loss of generality we may assume that ${\cal F}\subset A^{\textbf{1}}$ . Moreover, without loss of generality (omitting an error within $\varepsilon /16$ , say), we may further assume that there is $e_A\in A_+^{\textbf{1}}$ such that

(e3.1) $$ \begin{align} e_Ax=x=xe_A \mathrm{\ for\ all\ } x\in {\cal F}. \end{align} $$

Set $a\in A_+\setminus \{0\}$ , let $\varepsilon>0$ and let ${\cal G}=\{g_1,g_2,\dotsc ,g_n\}\subset C_0([0,1])$ be a finite subset.

By the Weierstrass theorem, there are $m\in \mathbb {N}$ and polynomials $p_i(t)=\sum _{k=1}^m \beta _k^{\left (i\right )}t^k$ such that

(e3.2) $$ \begin{align} \lvert p_i(t)-g_i(t)\rvert<\varepsilon/4 \mathrm{\ for\ all\ }t\in[0,1] \mathrm{\ and\ all\ } i\in\{1,2,\dotsc,n\}. \end{align} $$

Let $M=1+\max \left \{\left \lvert \beta _k^{\left (i\right )}\right \rvert :i=1,2,\dotsc ,n, \ k=1,2,\dotsc ,m\right \}$ and $\delta := \frac {\varepsilon }{32m^3M}$ .

Now, since A is e. tracially in ${\cal P}$ , there exist an element $e\in A_+^{\textbf{1}}$ and a nonzero $C^*$ -subalgebra $B\subset A$ such that B in $\mathcal {P}$ , and the following hold:

1. (1) $\ \lVert ex-xe\rVert <\delta \text { for all } x\in {\cal F}\cup \{e_A\}$ .

2. (2^′) $(1-e)x\in _{\delta } B$ and $\lVert (1-e)x\rVert \ge \lVert x\rVert -\delta $ for all $x\in {\cal F}\cup \{e_A\}$ .

3. (3) $\ e\lesssim a$ .

It remains to show that $g_i(1-e)x\in _{\varepsilon /2} B$ for all $x\in {\cal F}$ , $i=1,2,\dotsc ,n$ .

Claim: For all $x\in {\cal F}$ and all $k\in \{1,2,\dotsc ,m\}$ , we have $(1-e)^kx\in _{\frac {\varepsilon }{16mM}} B$ . In fact,

(e3.3) $$ \begin{align} {{(1-e)^kx\overset{\text{(e3.1)}}{=}(1-e)^ke_A^{k-1}x \overset{(1)}{\approx_{k^2\delta}}\overbrace{(1-e)e_A(1-e)e_A\dotsm (1-e)e_A}^{k-1}(1-e)x \overset{(2')}{\in_{k\delta}} B.}} \end{align} $$

Note that $2k^2\delta \le 2m^2\delta <\varepsilon /16mM$ . The claim follows.

By formula (e3.2) and the claim, for $x\in {\cal F}$ and $i\in \{1,2,\dotsc ,n\}$ we have

(e3.4) $$ \begin{align} g_i(1-e)x\approx_{\varepsilon/4}p_i(1-e)x =\sum_{k=1}^m{\beta}_k^{\left(i\right)} (1-e)^k x\in_{\varepsilon/4} B.\nonumber \\[-26pt] \end{align} $$

Remark 3.3.

1. (1) A similar notion as in Definition 3.1 could also be defined for nonsimple $C^*$ -algebras. However, in the present paper we are interested in only the simple case.

2. (2) Note that in Proposition 3.2, $g(1-e)$ is an element in ${\widetilde A}$ . But $g(1-e)x\in A$ . In the case that A is unital, the condition $\lVert (1-e)x\rVert \ge \lVert x\rVert -\varepsilon $ for all $x\in {\cal F}$ in condition (2) of the definition 3.1 is redundant for most cases (we leave the discussion to [22]).

3. (3) The notion of tracial approximation was first introduced in [29] (see also [30]). Let ${\cal P}$ be a class of unital $C^*$ -algebras – for example, the class of $C^*$ -algebras which are isomorphic to $C^*$ -algebras of the form $C([0,1], F)$ , where F are finite-dimensional $C^*$ -algebras. If, in Definition 3.1, $1-e$ can be chosen to be the unit of $B (\in {\cal P})$ , then A is TAI or A has tracial rank at most $1$ [30, 32]. In general, if A is unital simple and is TA ${\cal P}$ (see [14, Definition 2.2] and [18]), then A is e. tracially in ${\cal P}$ . The difference is that we allow e to be a positive element rather than a projection.

   To see this, let A be a unital simple $C^*$ -algebra which is TA ${\cal P}$ . Fix a finite subset ${\cal F}\subset A$ that contains $1_A$ . Fix $\varepsilon>0$ and $a\in A_+\backslash \{0\}$ . By a well-known result due to Blackadar (see, for example, [3, II.8.5.6]), there is a unital separable simple $C^*$ -subalgebra $C\subset A$ such that ${\cal F}\subset C$ . Let $\{{\cal F}_n\}_{n\in \mathbb {N}}$ be an increasing sequence of finite subsets of C whose union is dense in C, and ${\cal F} \subset {\cal F}_1$ . Since A is TA ${\cal P}$ in the sense of [14, Definition 2.2], there are nonzero projections $p_n\in A$ and $C^*$ -algebras $B_n\subset A$ with $B_n$ in ${\cal P}$ , and $p_n$ is the unit of $B_n$ ( $n\in \mathbb {N}$ ), which satisfies

   1. (i) $\lVert p_nx-xp_n\rVert <\varepsilon /2n$ for all $x\in {\cal F}_n$ ,

   2. (ii) $p_nxp_n\in _{\varepsilon /2n}B_n$ for all $x\in {\cal F}_n$ and

   3. (iii) $1-p_n\lesssim a$ .

   Assume that for each $n\in \mathbb {N}$ , there is some $x\in {\cal F}$ such that $\lVert p_nxp_n\rVert \le \lVert p_nx\rVert <\lVert x\rVert -\varepsilon $ . Then since ${\cal F}$ is a finite set, we can find $x_0\in {\cal F}$ and an increasing sequence of natural numbers $\{n_m\}_{m\in \mathbb {N}}$ such that $\left \lVert p_{n_m}x_0p_{n_m}\right \rVert <\lVert x_0\rVert -\varepsilon $ for all $m\in \mathbb {N}$ . Define a c.p.c. linear map $\varphi : C\to l^\infty (A)/c_0(A)$ by $\varphi (x):=\pi \left (\left \{p_{n_1}xp_{n_1},p_{n_2}xp_{n_2},\dotsc \right \}\right )$ , where $x\in C$ and $\pi :l^\infty (A)\to l^\infty (A)/c_0(A)$ is the quotient map. By condition (i) we see that $\varphi $ is a homomorphism. Since $\varphi (1_A)= \pi \left (\left \{p_{n_1},p_{n_2},\dotsc \right \}\right )\neq 0$ , $\varphi $ is nonzero. Since C is simple, $\varphi $ is injective and hence isometric. However, $\lVert \varphi (x_0)\rVert =\left \lVert \pi \left (\left \{p_{n_1}x_0,p_{n_2}x_0,\dotsc \right \}\right )\right \rVert \leq \sup _{m\in \mathbb {N}}\left \lVert p_{n_m}x_0p_{n_m}\right \rVert \leq \lVert x_0\rVert -\varepsilon $ : a contradiction. Therefore, there is $n_0\in \mathbb {N}$ such that $\left \lVert p_{n_0}x\right \rVert \geq \lVert x\rVert -\varepsilon $ for all $x\in {\cal F}$ . Set $e:=1_A-p_{n_0}$ ; then by (i)–(iii) and the choice of $n_0$ , we have

   1. (1^′) $\lVert ex-xe\rVert <\varepsilon $ for all $x\in {\cal F}$ ,

   2. (2^′) $(1-e)x\in _{\varepsilon }B_{n_0}$ and $\lVert (1-e)x\rVert \geq \lVert x\rVert -\varepsilon $ for all $x\in {\cal F}$ and

   3. (3^′) $e\lesssim a$ .

   Hence A is e. tracially in ${\cal P}$ .

   We note also that in general, a $C^*$ -algebra that is essentially tracially in ${\cal P}$ may not be TA ${\cal P}$ (see Remark 8.5).

4. (4) The current definition is also related to the notion of a ‘centrally large subalgebra’ ([37, Definition 4.1] and [1, Definition 2.1]) but not the same. The main difference is that the $C^*$ -subalgebra B in [1, Definition 2.1] is fixed. In fact, for a simple unital $C^*$ -algebra A and a class of $C^*$ -algebras ${\cal P}$ , if A has a centrally large subalgebra B with $B\in {\cal P}$ , then A is essentially tracially in ${\cal P}$ . On the other hand, in general, if A is essentially tracially in ${\cal P}$ , one may not find a centrally large $C^*$ -subalgebra B which is in ${\cal P}$ (for example, if ${\cal P}$ is the class of finite-dimensional $C^*$ -algebras, then every unital infinite-dimensional simple AF-algebra is e. tracially in ${\cal P}$ , but may not have centrally large finite-dimensional $C^*$ -subalgebras [37, Theorem 6.8]).

5. (5) In [21], a notion of asymptotically tracial approximation is introduced, studying tracial approximation of certain properties which are closely related to weakly stable relations. It also mainly studies unital simple $C^*$ -algebras with a rich structure of projections. This is different from Definition 3.1. However, if A is a unital (infinite-dimensional) simple $C^*$ -algebra which is asymptotically tracially in the class ${\cal C}$ of $1$ -dimensional noncommutative CW complexes, then one can show that A is also essentially tracially in the same class ${\cal C}$ . Moreover, many classes ${\cal P}$ of $C^*$ -algebras are preserved by asymptotically tracial approximation [21, Section 4]. Some more discussion may be found in a forthcoming paper [22].

Definition 3.4. Let ${\cal P}$ be a class of $C^*$ -algebras. The class ${\cal P}$ is said to have property (H) if, for any nonzero A in ${\cal P}$ and any nonzero hereditary $C^*$ -subalgebra $B\subset A$ , B is also in ${\cal P}$ .

Proposition 3.5. Let ${\cal P}$ be a class of $C^*$ -algebras which has property (H). Suppose that A is a simple $C^*$ -algebra which is e. tracially in ${\cal P}$ . Then every nonzero hereditary $C^*$ -subalgebra $B\subset A$ is also e. tracially in ${\cal P}$ .

Proof. Assume ${\cal P}$ has property (H) and A is e. tracially in ${\cal P}$ . Let $B\subset A$ be a nonzero hereditary $C^*$ -subalgebra of A. Set ${\cal F}\subset B$ and $s\in B_+\setminus \{0\}$ , and $\varepsilon \in (0,1/4)$ .

Without loss of generality, we may assume that ${\cal F}\subset B_+^{\textbf{1}}$ . Let $d\in B_+^{\textbf{1}}$ be such that $dx\approx _{{\varepsilon /32}} x\approx _{{\varepsilon /32}} xd$ and $x\approx _{{{\varepsilon /32}}}dxd$ for all $x\in {\cal F}$ .

Put $\varepsilon _1=\varepsilon /32$ . By [15, Lemma 3.3], there is $\delta _1\in (0,{{\varepsilon _1}})$ such that for any $C^*$ -algebra E and any $x,y\in E_+^{\textbf{1}}$ , if $x\approx _{\delta _1} y$ , then there is an injective homomorphism $\psi : \mathrm {Her}_{E}\left (f_{{{\varepsilon _1}}/2}(x)\right )\to \mathrm {Her}_{E}(y)$ satisfying $z\approx _{{\varepsilon _1}} \psi (z)$ for all $z\in \mathrm {Her}_{E}\left (f_{{{\varepsilon _1}}/2}(x)\right )^{\textbf{1}}$ .

Note that there is $\delta _2\in (0,\delta _1)$ such that for any $C^*$ -algebra E and any $x,y\in E_+^{\textbf{1}}$ , if $xy\approx _{\delta _2} yx$ , then $x^{1/4}y\approx _{\delta _1/2} yx^{1/4}$ , $x^{1/8}y^{1/2}\approx _{\delta _1/2} y^{1/2}x^{1/8}$ and $x^{1/8}y\approx _{\delta _1/2 }yx^{1/8}$ .

Let $\delta =\delta _2/2$ . Let ${\cal G}=\left \{t,t^{1/4},t^{1/8}\right \}\subset C_0([0,1])$ . Since A is e. tracially in ${\cal P}$ , by Proposition 3.2 there exist a positive element $a\in A_+^{\textbf{1}}$ and a nonzero $C^*$ -subalgebra $C\subset A$ which is in ${\cal P}$ such that

1. (1) $\lVert ax-xa\rVert <{\delta }$ for all $x\in {\cal F}\cup \left \{d,d^{1/2},d^2\right \}$ ,

2. (2) $g(1-a) x\in _{\delta } C$ for all $g\in {\cal G}$ and $\lVert (1-a)x\rVert \ge \lVert x\rVert -\delta $ for all $x\in {\cal F}\cup \left \{d,d^{1/2},d^2\right \}$ and

3. (3) $a\lesssim s$ .

By (2), there is $c\in C$ such that $ c\approx _{\delta _1/2} (1-a)^{1/4}d$ . By (1) and the choice of $\delta _2$ , we have $c\approx _{\delta _1}d^{1/2}(1-a)^{1/4}d^{1/2}$ . Then by [15, Lemma 3.3] and the choice of $\delta _1$ , there is a monomorphism

$$ \begin{align*} \varphi: \mathrm{Her}_A\left(f_{{{\varepsilon_1}}/2}(c)\right)\to \mathrm{Her}_A\left(d^{1/2}(1-a)^{1/4}d^{1/2}\right)\subset B \end{align*} $$

satisfying $\lVert \varphi (x)-x\rVert <{{\varepsilon _1}}$ for all $x\in \mathrm {Her}_C\!\left(f_{{{\varepsilon _1}}/2}(c)\right )^{\textbf{1}}$ . Define $D:=\varphi \left (\mathrm {Her}_C\!\left(f_{{{\varepsilon _1}}/2}(c)\right )\right )\subset B$ . Since C is in ${\cal P}$ and ${\cal P}$ has property (H), $D\cong \mathrm {Her}_C\!\left(f_{\varepsilon _1/2}(c)\right )$ is in ${\cal P}$ . Set $b:=dad\in B_+^{\textbf{1}}$ . Then by (1) and the choice of d, we have

(e3.5) $$ \begin{align} \lVert bx-xb\rVert=\lVert dadx-xdad\rVert \approx_{4{{\varepsilon_1}}} \lVert adxd-dxda\rVert \approx_{2{{\varepsilon_1}}} \lVert ax-xa\rVert<\delta \mathrm{\ for\ all\ }x\in{\cal F}. \end{align} $$

By (2), for any $x\in {\cal F}$ there is $\bar x\in C$ such that $(1-a)^{1/4}x(1-a)^{1/4}\approx _{2{{\varepsilon _1}}} \bar x$ . Then

(e3.6) $$ \begin{align} \begin{split} (1-b)x &= (1-dad)x \approx_{3{{\varepsilon_1}}} (1-a)dxd\\ &\approx_{4{{\varepsilon_1}}} (1-a)^{1/8}d(1-a)^{1/8} \cdot(1-a)^{1/4}x(1-a)^{1/4}\cdot (1-a)^{1/8}d(1-a)^{1/8}\\ &\approx_{4{{\varepsilon_1}}} c\bar xc \approx_{2{{\varepsilon_1}}} (c-{{\varepsilon_1}})_+\bar x (c-{{\varepsilon_1}})_+\\ &\approx_{{{\varepsilon_1}}} \varphi((c-{{\varepsilon_1}})_+\bar x (c-{{\varepsilon_1}})_+)\in D. \end{split} \end{align} $$

In other words,

(e3.7) $$ \begin{align} (1-b)x\in_{\varepsilon} D. \end{align} $$

Therefore, for all $x\in {\cal F}$ ,

(e3.8) $$ \begin{align} \begin{split} \lVert(1-b)x\rVert&=\lVert(1-dad)x\rVert \ge \left\lVert\left(1-a d^2\right)x\right\rVert-\delta \\ &\ge \lVert(1-a)x\rVert-3\varepsilon_1 \ge \lVert x\rVert-\delta-3\varepsilon_1\ge \lVert x\rVert-\varepsilon. \end{split} \end{align} $$

By (3), we have $b=dad\lesssim _A s$ . Note that $b, s\in B$ . Since B is a hereditary $C^*$ -subalgebra, we have $b\lesssim _B s$ . By formulas (e3.5) and (e3.7), we see that B is also e. tracially in ${\cal P}$ .

4. Basic properties

Notation 4.1. Let ${\cal W}$ be the class of $C^*$ -algebras A such that $W(A)$ is almost unperforated.

Let ${\mathscr Z}$ be the Jiang–Su algebra [27]. A $C^*$ -algebra A is called ${\mathscr Z}$ -stable if $A\otimes {\mathscr Z}\cong A$ . Let ${\cal C}_{\mathscr Z}$ be the class of separable ${\mathscr Z}$ -stable $C^*$ -algebras.

Lemma 4.2. Let A be a simple $C^*$ -algebra which is e. tracially in ${\cal W}$ , and set $a,b,c\in A_+\backslash \{0\}$ . Suppose that there exists $n\in \mathbb {N}$ satisfying $(n+1)\langle a\rangle \leq n\langle b\rangle $ . Then for any $\varepsilon>0$ , there exist $a_1,a_2\in A_+$ such that

1. (1) $a\approx _{\epsilon }a_1+a_2$ ,

2. (2) $a_1\lesssim _A b$ and

3. (3) $a_2 \lesssim _A c$ .

Proof. Without loss of generality, one may assume that $a,b,c\in A_+^{\textbf{1}}\backslash \{0\}$ and $\epsilon <1/2$ . Then $(n+1)\langle a\rangle \leq n\langle b\rangle $ implies that there exists $r=\sum _{i,j=1}^{n+1}r_{i,j}\otimes e_{i,j}\in A\otimes M_{n+1}$ such that

(e4.1) $$ \begin{align} a\otimes \sum_{i=1}^{n+1}e_{i,i} \approx_{\epsilon/128}r^*\left(b \otimes \sum_{i=1}^{{{n}}}e_{i,i}\right)r. \end{align} $$

Set ${\cal F}:=\{a,b\}\cup \left \{r_{i,j},r_{i,j}^*:i,j=1, 2 ,\dotsc , n+1\right \}$ and $M:=1+\lVert r\rVert $ . Let $\sigma =\frac {\varepsilon }{32M^2(n+1)^4}$ . Since A is e. tracially in ${\cal W}$ , by Proposition 3.2, for any $\delta \in \left (0, \frac {\varepsilon }{256M(n+1)^2}\right )$ , there exist $f\in A_+^{\textbf{1}}\setminus \{0\}$ and a $C^*$ -subalgebra $B\subset A$ which has almost unperforated $W(B)$ such that

1. (1^′) $\lVert fx-xf \rVert <\delta $ for $x\in {\cal F}$ ,

2. (2^′) $(1-f)^{ 1/4} x, (1-f)^{ 1/2} a(1-f)^{ 1/2 }, (1-f)^{1/4}x (1-f)^{ 1/4} \in _\delta B$ for all $x\in {\cal F}$ and

3. (3^′) $f\lesssim c$ .

Put $g=1-f$ . Let

$$ \begin{align*} {\cal G}:=\left\{g^{ {1/4}}x, g^{1/2}xg^{1/2}, g^{1/4}xg^{1/4} : x\in {\cal F}\right\}. \end{align*} $$

Set $x\in {\cal G}\cap A_+$ . By (2 $'$ ), there is $\bar x\in B$ such that $\lVert x-\bar x\rVert <\delta $ . Let $x':=\left (\bar x+\bar x^*\right )/2\in B_{\mathrm {sa}}$ . Then $x\approx _\delta x'$ . Then $x'+\delta \geq x\geq 0$ , which implies $\left \lVert x^{\prime }_-\right \rVert \leq \delta $ . Then $x\approx _\delta x'=x^{\prime }_+-x^{\prime }_-\approx _\delta x^{\prime }_+\in B_+$ . Therefore, there is a map ${\alpha }:{\cal G}\to B$ such that ${\alpha }({\cal G}\cap A_+)\subset B_+$ , and

(e4.2) $$ \begin{align} x \approx_{2\delta} {\alpha}( x ) \mathrm{\ for\ all\ }x\in{\cal G}. \end{align} $$

From (1 $'$ ) and (2 $'$ ), one can choose $\delta $ sufficiently small such that

(e4.3) $$ \begin{align} a\approx_{\varepsilon/16} g^{{1/2}}ag^{{1/2}}+ (1-g)^{1/2}a(1-g)^{1/2} \mathrm{\ and\ } \end{align} $$

(e4.4) $$ \begin{align} \left(g^{{1/2}}{{a}}g^{{1/2}}-\varepsilon/8\right)_+ \approx_{\varepsilon/16} \left(\alpha\left(g^{{1/2}}{{a}} g^{{1/2}}\right)-\epsilon/8\right)_+. \end{align} $$

By (1 $'$ ) and formula (e4.1) (with $\delta $ sufficiently small), one can also assume that

(e4.5) $$ \begin{align} g^{{1/2}}ag^{{1/2}}\otimes \sum_{i=1}^{n+1}e_{i,i} \approx_{\varepsilon/64} R^* \left(g^{{1/4}} b g^{{1/4}}\otimes \sum_{i=1}^{n}e_{i,i}\right)R, \end{align} $$

where $R:=\sum _{i,j=1}^{n+1}\left ({{g^{1/4}}}{{r_{i,j}}} \right )\otimes e_{i,j}$ . By formulas (e4.5) and (e4.2) and $\delta <\frac {\varepsilon }{256M(n+1)^2}$ , one has

(e4.6) $$ \begin{align} \alpha\left(g^{{1/2}}{{a}}g^{{1/2}}\right) \otimes \sum_{i=1}^{n+1}e_{i,i} \approx_{\varepsilon/32} {{\bar R^*}} \left( {\alpha}\left(g^{1/4}{{b}}g^{1/4}\right) \otimes \sum_{i=1}^{n}e_{i,i}\right){{\bar R}}, \end{align} $$

where $\bar R:= \sum _{i,j=1}^{n+1}{\alpha }\left (g^{1/4}{{r_{i,j}}} \right )\otimes e_{i,j}$ . Then by the choice of $\sigma $ ,

(e4.7) $$ \begin{align} \alpha\left(g^{1/2}{a}g^{1/2}\right) \otimes \sum_{i=1}^{n+1}e_{i,i} \approx_{\varepsilon/16} {{\bar R^*}} \left( \left({\alpha}\left(g^{1/4}{{b}}g^{1/4}\right)-\sigma\right)_+ \otimes \sum_{i=1}^{n}e_{i,i}\right){{\bar R}}. \end{align} $$

By formula (e4.7) and [40, Proposition 2.2], one has

(e4.8) $$ \begin{align} \left(\alpha\left(g^{{1/2}}{{a}}g^{{1/2}}\right) -\epsilon/8\right)_+\otimes \sum_{i=1}^{n+1}e_{i,i} \lesssim \left({\alpha}\left(g^{1/4} b g^{1/4}\right) -\sigma\right)_+\otimes \sum_{i=1}^{n}e_{i,i}. \end{align} $$

Since $W(B)$ is almost unperforated, one obtains

(e4.9) $$ \begin{align} \left(\alpha\left(g^{{1/2}}{{a}}g^{{1/2}}\right) -\epsilon/8\right)_+ \lesssim \left({\alpha}\left(g^{1/4}{{b}}g^{1/4}\right) -\sigma\right)_+. \end{align} $$

By [40, Proposition 2.2] and formulas (e4.4), (e4.9) and (e4.2), it follows that

(e4.10) $$ \begin{align} \left(g^{{1/2}}{{a}}g^{{1/2}}-\varepsilon/4\right)_+ &\lesssim \left(\alpha\left(g^{{1/2}}{{a}}g^{{1/2}}\right) -\epsilon/8\right)_+ \hspace{6pc} \end{align} $$

(e4.11) $$ \begin{align} & \hspace{6pc} \lesssim \left({\alpha}\left(g^{1/4}{{b}}g^{1/4}\right) -\sigma\right)_+ \lesssim g^{1/4}{{b}}g^{1/4} \lesssim b. \end{align} $$

By (1 $'$ ) and the choice of $\delta $ ,

(e4.12) $$ \begin{align} a\approx_{\varepsilon/16} (1-f)^{1/2}a(1-f)^{1/2}+f^{1/2} a f^{1/2}. \end{align} $$

Choose

(e4.13) $$ \begin{align} a_1&:=\left(g^{{1/2}}a g^{{1/2}}-\varepsilon/2\right)_+ =\left((1-f)^{1/2}a(1-f)^{1/2}-\varepsilon/2\right)_+ \mathrm{\ and\ } \end{align} $$

(e4.14) $$ \begin{align} a_2&:=f^{1/2}af^{1/2}. \hspace{16pc} \end{align} $$

Then by formula (e4.11), one has $a_1\lesssim _A b$ . Note that (3 $'$ ) implies $a_2 \lesssim _A c$ . Thus $a_1$ and $a_2$ satisfy (2) and (3) of the lemma. By formula (e4.12),

$$ \begin{align*} a\approx_{\varepsilon/16} (1-f)^{{1/2}}a(1-f)^{{1/2}}+f^{{1/2}}af^{{1/2}} \approx_{\epsilon/2} a_1+a_2. \end{align*} $$

So (1) of the lemma also holds, and the lemma follows.

Theorem 4.3. Let A be a simple $C^*$ -algebra which is e. tracially in ${\cal W}$ (see Notation 4.1). Then $A\in {\cal W}$ .

Proof. We may assume that A is nonelementary. Set $a,b\in M_m(A)_+\setminus \{0\}$ with $\lVert a\rVert =1=\lVert b\rVert $ for some integer $m\ge 1$ . Set $n\in \mathbb {N}$ and assume $(n+1)\langle a\rangle \leq n\langle b\rangle $ . To prove the theorem, it suffices to prove that $a\lesssim b$ .

Note that if $B\in {\cal W}$ , then for each integer m, we have $M_m(B)\in {\cal W}$ . It follows that $M_m(A)$ is e. tracially in ${\cal W}$ . To simplify notation, without loss of generality one may assume $a, b\in A_+$ .

By [21, Lemma 4.3], $\mathrm {Her}\left (f_{1/4}(b)\right )_+$ contains $2n+1$ nonzero mutually orthogonal elements $b_0, b_1, \dotsc ,b_{2n}$ such that $\langle b_i\rangle =\langle b_0\rangle $ , $i=1, 2, \dotsc ,2n$ . Without loss of generality, we may assume that $\lVert b_0\rVert =1$ . If $b_0$ is a projection, choose $e_0=b_0$ . Otherwise, by replacing $b_0$ by $g_1(b_0)$ for some continuous function $g_1\in C_0([0,1])$ , we may assume that there is a nonzero $e_0\in A_+$ such that $b_0e_0=e_0b_0=e_0$ . Replacing b by $g(b)$ for some $g\in C_0((0, 1])$ , one may assume that $bb_0=b_0b=b_0$ . Put $c=b-b_0$ . Note that

(e4.15) $$ \begin{align} ce:0=(b-b_0)e_0=be_0-e_0=b_0e_0-e_0=0=e_0c. \end{align} $$

Keep in mind that $b\ge c + e_0$ , $c \perp e_0$ and $2n\langle b_0\rangle \le {{\langle c\rangle =\langle b-b_0 \rangle }}$ . One has

(e4.16) $$ \begin{align} (2n+2)\langle a\rangle \le 2n \langle b\rangle \le 2n(\langle b-b_0\rangle +\langle b_0\rangle)\le 2n\langle c\rangle +\langle c\rangle=(2n+1)\langle c \rangle. \end{align} $$

By Lemma 4.2, for any $\varepsilon \in (0,1/2)$ there exist $a_1,a_2\in A_+$ such that

1. (i) $a\approx _{\epsilon /2}a_1+a_2$ ,

2. (ii) $a_1\lesssim _A c$ and

3. (iii) $a_2\lesssim _A e_0$ .

By (i)–(iii) and applying [40, Proposition 2.2] (recall $be_0=e_0b=e_0$ ), one obtains

(e4.17) $$ \begin{align} (a-\varepsilon)_+\lesssim a_1+a_2 \lesssim c+e_0\le b. \end{align} $$

Since this holds for every $\varepsilon \in (0,1/2)$ , one concludes that $a\lesssim b$ .

Corollary 4.4. Let A be a simple $C^*$ -algebra which is e. tracially in ${\cal C}_{\mathscr Z}$ . Then $W(A)$ is almost unperforated.

Proof. It follows from [42, Theorem 4.5] and Theorem 4.3.

Definition 4.5. Let A be a $C^*$ -algebra. Let ${\cal T}$ denote the class of $C^*$ -algebras A such that for every $a\in \mathrm {Ped}(A)_+\setminus \{0\}$ , every $2$ -quasitrace of $\overline {aAa}$ is a trace.

Set $A\in {\cal T}$ and let $B\subset A$ be a hereditary $C^*$ -subalgebra. If $b\in \mathrm {Ped}(B)_+\setminus \{0\}$ , then $b\in \mathrm {Ped}(A)_+$ and $\overline {bBb}=\overline {bAb}$ . It follows that every $2$ -quasitrace of $\overline {bBb}$ is a trace. Hence ${\cal T}$ has property (H).

Proposition 4.6. Let A be a simple $C^*$ -algebra which is e. tracially in ${\cal T}$ . Then A is in ${\cal T}$ .

Proof. Fix $a\in \mathrm {Ped} (A)_+^{\textbf{1}}$ and let $C=\mathrm {Her}(a)$ . We will show that every $2$ -quasitrace of C is a trace. We may assume that C is nonelementary. Set $\tau \in QT(C)$ . Fix $x, y\in C_{\mathrm {sa}}$ with $\lVert x\rVert ,\lVert y\rVert \leq 1/2$ . Set $\varepsilon \in (0,1/2)$ . Let ${\cal F}:=\{x,y,x+y\}$ . Let $n\in \mathbb {N}$ be such that $\varepsilon>1/n$ . By [21, Lemma 4.3], there exist mutually orthogonal norm $1$ positive elements $c_1,c_2,\dotsc , c_n\in A_+\backslash \{0\}$ such that $c_1\sim c_2\sim \dotsb \sim c_n$ . Then $d_{\tau } ( c_1 )\leq 1/n<\varepsilon $ .

Let $\delta \in (0,\varepsilon )$ be such that for any $d\in C_+^{\textbf{1}}$ and $z\in C_{\mathrm {sa}}^{\textbf{1}}$ , if $\lVert [d,z]\rVert <\delta $ , then

(e4.18) $$ \begin{align} z\approx_\varepsilon (1-d)^{1/2}z(1-d)^{1/2}+d^{1/2}zd^{1/2} \end{align} $$

and

(e4.19) $$ \begin{align} \tau(z)\approx_\varepsilon \tau\left((1-d)^{1/2}z(1-d)^{1/2}\right)+\tau\left(d^{1/2}zd^{1/2}\right). \end{align} $$

(Note [4, II.2.6] that $\left \lVert \left [(1-d)^{1/2}z(1-d)^{1/2},d^{1/2}zd^{1/2}\right ]\right \rVert $ can be sufficiently small depending on $\delta $ .) Note that ${\cal T}$ has property (H). Since A is simple and e. tracially in ${\cal T}$ , by Proposition 3.5 C is also e. tracially in ${\cal T}$ . There exist an element $e\in C_+^{\textbf{1}}$ and a nonzero $C^*$ -subalgebra $B\subset C$ such that B is in $\mathcal {T}$ , and the following are true:

1. (1) $\lVert ez-ze\rVert <\delta $ for all $z\in \cal F$ .

2. (2) $(1-e)^{1/2}z(1-e)^{1/2}\in _{\delta /2} B$ for all $z\in {\cal F}$ .

3. (3) $e\lesssim c_1$ .

We may choose $e_B\in \mathrm {Ped}(B)_+^{\textbf{1}}$ such that

1. (2^′) $(1-e)^{1/2}z(1-e)^{1/2}\in _{\delta } B_1:=\overline {e_BBe_B}$ for all $z\in {\cal F}$ .

Note that for $z\in {\cal F}$ , $e^{1/2}ze^{1/2}$ is self-adjoint. One has $\left (e^{1/2}ze^{1/2}\right )_+,\left (e^{1/2}ze^{1/2}\right )_-\in \mathrm {Her}_A(e)$ . Then

(e4.20) $$ \begin{align} \left\lvert\tau\left(e^{1/2}ze^{1/2}\right)\right\rvert &= \left\lvert\tau\left(\left(e^{1/2}ze^{1/2}\right)_+\right)-\tau\left(\left(e^{1/2}ze^{1/2}\right)_-\right)\right\rvert \hspace{6pc} \end{align} $$

(e4.21) $$ \begin{align} & \hspace{7pc} \leq d_\tau\left(\left(e^{1/2}ze^{1/2}\right)_+\right)+d_\tau\left(\left(e^{1/2}ze^{1/2}\right)_-\right) \leq 2d_\tau(e)\leq 2\varepsilon. \end{align} $$

Then by (1), the choice of $\delta $ and formulas (e4.18) and (e4.19), for $z\in {\cal F}$ ,

(e4.22) $$ \begin{align} \hspace{3pc} \tau(z) &\approx_{2\varepsilon} \tau\left((1-e)^{1/2}z(1-e)^{1/2}\right)+\tau\left(e^{1/2}ze^{1/2}\right) \end{align} $$

(e4.23) $$ \begin{align} \text{(by formula (e4.21))}&\approx_{2\varepsilon} \tau\left((1-e)^{1/2}z(1-e)^{1/2}\right). \hspace{7pc} \end{align} $$

By (2 $'$ ), there are $\bar x, \bar y\in (B_1)_{\mathrm {sa}}$ such that

(e4.24) $$ \begin{align} (1-e)^{1/2}x(1-e)^{1/2}\approx_{2\delta} \bar x, \qquad (1-e)^{1/2}y(1-e)^{1/2}\approx_{2\delta} \bar y. \end{align} $$

Then

$$ \begin{align*} \tau(x+y) &\overset{\text{formula (e 4.23)}}{\approx_{{{4\varepsilon}}}} \tau\left((1-e)^{1/2}(x+y)(1-e)^{1/2}\right) \\ &\overset{\text{formula (e 4.24)}}{\approx_{{4\delta}}} \tau\left(\bar x+\bar y\right) \\ \text{(}\tau\ \text{is a trace on}\ B_1\text{)}&\overset{\phantom{\text{formula (e 4.24)}}}{=} \tau(\bar x)+\tau(\bar y) \\ &\overset{\text{formula (e 4.24)}}{\approx_{{4\delta}}} \tau\left((1-e)^{1/2}x(1-e)^{1/2}\right)+ \tau\left((1-e)^{1/2}y(1-e)^{1/2}\right) \\ &\overset{\text{formula (e 4.23)}}{\approx_{{4\varepsilon}}} \tau(x)+ \tau(y). \end{align*} $$

Since $\varepsilon $ and $\delta $ are arbitrary small, we have $\tau (x+y)=\tau (x)+\tau (y)$ , and therefore $\tau $ is a trace on C.

Definition 4.7. Let A be a $C^*$ -algebra. Recall that an element $a\in \mathrm {Ped}(A)_+$ is said to be infinite if there are nonzero elements $b, c\in \mathrm {Ped}(A)_+$ such that $bc=cb=0$ , $b+c\lesssim c$ and $c\lesssim a$ . A is said to be finite if every element $a\in \mathrm {Ped}(A)_+$ is not infinite (see, for example, [33, Definition 1.1]). A is stably finite if $M_n(A)$ is finite for every integer $n\ge 1$ .

Recall that a simple $C^*$ -algebra A is purely infinite if and only if every nonzero element in $\mathrm {Ped}(A)_+$ is infinite (see [33, Condition (vii), Theorem 2.2]). Let ${\cal PI}$ be the class of $C^*$ -algebras such that every nonzero positive element in the Pedersen ideal is infinite.

Theorem 4.8. Let A be a simple $C^*$ -algebra which is e. tracially in ${\cal PI}$ . Then A is purely infinite.

Proof. Note that A has infinite dimension. Set $a\in \mathrm {Ped}(A)_+\setminus \{0\}$ with $\lVert a\rVert =1$ .

Since $\overline {f_{1/4}(a)Af_{1/4}(a)}$ is an infinite-dimensional simple $C^*$ -algebra, one may choose $c, d\in \overline {f_{1/4}(a)Af_{1/4}(a)}_+\setminus \{0\}$ such that $cd=dc=0$ .

Since A is e. tracially in ${\cal PI}$ , there exist a sequence of positive elements $e_n\in A_+$ with $\lVert e_n\rVert \le 1$ and a sequence of $C^*$ -subalgebra $B_n\subset A$ such that $B_n$ in ${\cal PI}$ , and the following are true:

1. (1) $a\approx _{1/2^n} e_n^{1/2}ae_n^{1/2}+(1-e_n)^{1/2}a(1-e_n)^{1/2}$ .

2. (2) $(1-e_n)^{1/2}a(1-e_n)^{1/2}\in _{1/2^n} B_n$ and $\left \lVert (1-e_n)^{1/2}a(1-e_n)^{1/2}\right \rVert \ge \lVert a\rVert -{1/2^n}$ .

3. (3) $e_n\lesssim c$ .

By (2), there is $b_n\in {B_n}_+$ such that $b_n\approx _{1/2^n} (1-e_n)^{1/2}a(1-e_n)^{1/2}$ . Then by (1),

(e4.25) $$ \begin{align} a\approx_{2/2^n} b_n+e_n^{1/2}ae_n^{1/2}. \end{align} $$

Note that $\inf _n\{\lVert b_n\rVert \}\ge \lVert a\rVert /2>0$ . Choose $0<\varepsilon <\lVert a\rVert /16$ .

By [37, Lemma 1.7], for all sufficiently large n we have

(e4.26) $$ \begin{align} {{0\neq(b_{n}-2\varepsilon)_+\lesssim \left(b_{n}+e_{n}^{1/2}ae_{n}^{1/2}-2\varepsilon\right)_+\lesssim a.}} \end{align} $$

Note that $(b_{n}-2\varepsilon )_+\in \mathrm {Ped}(B_n)_+\setminus \{0\}$ . Then there are $d_1, d_2\in \mathrm {Ped}(B_n)_+\setminus \{0\}$ such that $d_1\perp d_2$ , $d_1+d_2\lesssim d_2\lesssim (b_{n}-2\varepsilon )_+$ and

(e4.27) $$ \begin{align} d_1+d_2\lesssim (b_{n}-2\varepsilon)_+\lesssim a. \end{align} $$

It follows that a is infinite, and therefore A is purely infinite.

Proposition 4.9 [42, Corollary 5.1]

Let A be a $\sigma $ -unital simple $C^*$ -algebrasuch that $W(A)$ is almost unperforated. If A is not purely infinite, then $\overline {aAa}$ has a nonzero $2$ -quasitrace for every $a\in \mathrm {Ped}(A)_+\setminus \{0\}$ . Consequently, A is stably finite.

Proof. This is a theorem of Rørdam [42, Corollary 5.1]. Since we do not assume that A is exact and will use only $2$ -quasitraces, some more explanation is in order. The explanation, of course, follows exactly the same lines as the proof of [42, Corollary 5.1].

Set $a\in \mathrm {Ped}(A)_+^{\textbf{1}}$ and $B:=\overline {aAa}$ . Then B is algebraically simple (see, for example, [3, II.5.4.2]). Assume that B has no nonzero $2$ -quasitraces.

Consider $W(B)$ . Note that $W(B)\subset W(A)$ , and $W(B)$ has the property that if $x\in W(B)$ and $y\in W(A)$ such that $y\le x$ , then $y\in W(B)$ . It follows that $W(B)$ is almost unperforated. Since B is algebraically simple, every element in $W(B)$ is a strong order unit.

Set $t, t'\in W(B)$ (with t a strong order unit). The statement (and the proof) of [40, Proposition 3.1] imply that if there is no state on $W(B)$ (with the strong order unit t), then there must be some integer $n\in \mathbb {N}$ and $u\in W(B)$ such that

(e4.28) $$ \begin{align} n t' +u\le nt +u. \end{align} $$

Then by [40, Proposition 3.2] (see the proof also), as $W(B)$ is almost unperforated,

(e4.29) $$ \begin{align} t'{{~\leq~}}t. \end{align} $$

On the other hand, by [4, II.2.2], every lower semicontinuous dimension function on $W(B)$ is induced by a $2$ -quasitrace on B. Since B is assumed to have no nonzero $2$ -quasitraces, combining with [40, Proposition 4.1] (as well as the paragraph before it) shows that there is no state on $W(B)$ . Therefore formula (e4.29) implies that for any $b, c\in B_+\setminus \{0\}$ , we have $b\lesssim c$ . It follows that B is purely infinite and so is A.

To see the last part of the statement, suppose that there are $b, c\in \mathrm {Ped}(A)_+^{\textbf{1}}\setminus \{0\}$ such that $bc=cb=0$ and $b+c\lesssim c$ . Let $a=b+c$ and $B=\overline {aAa}$ . Note that $a\in \mathrm {Ped}(A)_+$ . Then B has nonzero $2$ -quasitraces.

Therefore

(e4.30) $$ \begin{align} d_\tau(c)\ge d_\tau(b+c)\mathrm{\ for\ all\ } \tau\in QT(B). \end{align} $$

On the other hand, for any $\tau \in QT(B)$ and any $1>\varepsilon >0$ ,

(e4.31) $$ \begin{align} \tau(f_{\varepsilon}(b+c))=\tau(f_{\varepsilon}(b)+f_{\varepsilon}(c))=\tau(f_{\varepsilon}(b))+\tau(f_{\varepsilon}(c)). \end{align} $$

Fix $1>\varepsilon _0>0$ such that $f_{\varepsilon _0}(b)\not =0$ . Since B is algebraically simple, $\tau \left (f_{\varepsilon _0}(b)\right )>0$ for all $2$ -quasitraces $\tau $ . Fix $\tau \in QT(B)$ . Then, by equation (e4.31),

(e4.32) $$ \begin{align} d_\tau(b+c)\ge \tau\left(f_{\varepsilon_0}(b)\right)+d_\tau(c)>d_\tau(c). \end{align} $$

This contradicts formula (e4.30). It follows that no such pairs b and c exist. Thus A is finite.

Since $M_n(A)$ has the same relevant property as A, we conclude that A is stably finite.

Corollary 4.10. Let A be a $\sigma $ -unital simple $C^*$ -algebrasuch that A is e. tracially in ${\cal W}$ . Then A has strict comparison.

Proof. By Theorem 4.3, $W(A)$ is almost unperforated. It follows from Remark 2.5 that $\mathrm {Cu}(A)$ is almost unperforated. Fix $e\in \mathrm {Ped}(A)_+\setminus \{0\}$ and let $B:=\mathrm {Her}(e)$ . As in the proof of Proposition 4.9, every lower semicontinuous dimension function on $W(B)$ is induced by a $2$ -quasitrace of B. Set $a, b\in (A\otimes {\cal K})_+$ such that $d_\tau (a)<d_\tau (b)$ for all $\tau \in QT(B)$ . By [17, Propositions 4.2, 4.6], $a\lesssim b$ .

5. Essentially tracially ${\mathscr Z}$ -stable $C^*$ -algebras

Recall from Notation 4.1 that ${\cal C}_{\mathscr Z}$ is the class of separable ${\mathscr Z}$ -stable $C^*$ -algebras.

Theorem 5.1. Let A be a $\sigma $ -unital simple $C^*$ -algebra which is e. tracially in ${\cal C}_{\mathscr Z}$ . Then A is either purely infinite or stably finite. Moreover, if A is not purely infinite, then it has strict comparison for positive elements.

Proof. It follows from [42, Theorem 4.5] that every $C^*$ -algebra B in ${\cal C}_{\mathscr Z}$ has almost unperforated $W(B)$ . It follows from Theorem 4.3 and Remark 2.5 that $\mathrm {Cu}(A)$ is almost unperforated. By Proposition 4.9, if A is not purely infinite, then it is stably finite, and by the proof of Corollary 4.10, A has strict comparison for positive elements.

Definition 5.2. Let A be a simple $C^*$ -algebra. A is said to be tracially approximately divisible if for any $\varepsilon>0$ , any ${\cal F}=\{x_1,x_2,\dotsc ,x_m\}\subset A$ , any element $e_F\in A_+^{\textbf{1}}$ with $e_Fy_i=y_i=y_ie_F$ for some $y_i\approx _{\varepsilon /4} x_i$ , $1\le i\le m$ , any $s\in A_+\setminus \{0\}$ , and any integer $n\ge 1$ , there are $\theta \in A_+^{\textbf{1}}$ , a $C^*$ -subalgebra $D\otimes M_n\subset A$ and a c.p.c. map ${\beta }: A\to A$ such that the following are true:

1. (1) $x\approx _\varepsilon x'+{\beta }(x)$ for all $x\in {\cal F}$ , where $\lVert x'\rVert \le \lVert x\rVert $ , $x'\in \mathrm {Her}(\theta )$ .

2. (2) ${\beta }(x)\in _\varepsilon D\otimes 1_n$ and $e_F{\beta }(x)\approx _{\varepsilon } {\beta }(x)\approx _\varepsilon {\beta }(x)e_F$ for all $x\in {\cal F}$ .

3. (3) $\theta \lesssim s. $

The notion of approximate divisibility for $C^*$ -algebras was introduced in [6]. The term ‘tracially approximate divisibility’ appeared in [32] (for special cases, see [32, Definition 5.3, proof of Theorem 5.4], [29, Lemma 6.10], [15, Definition 10.1]).

1. (1) If A is a unital separable simple $C^*$ -algebra which is approximately divisible, then it is tracially approximately divisible. To see this, recall that by [6, Theorem 1.4(d)], A has strict comparison. Let $\varepsilon>0$ , a finite subset ${\cal F}\subset A$ , $a\in A_+\setminus \{0\}$ and $n\in \mathbb {N}$ be given. We assume that $1\in {\cal F}$ . Choose an integer m such that $d_\tau (a)\ge 1/m$ for all $\tau \in QT(A)$ (recall that A is a unital separable simple $C^*$ -algebra and $QT(A)$ is a simplex, by [5, II.4.4]). Choose an integer $k>mn$ . It follows from [6, Corollary 2.10] that we may assume that ${\cal F}\subset _{\varepsilon /2} \bigoplus _{i=1}^s A_n\otimes 1_{M_{k_i}}\subset \bigoplus _{i=1}^s A_n\otimes M_{k_i}$ , where $A_n$ is a $C^*$ -subalgebra of A and $k_i\ge k$ . Write $k_i=l_imn+r_i$ , where $l_i, r_i\in \mathbb {N}$ and $0\le r_i < mn$ , $i=1,2,\dotsc ,s$ . Note that $1\in \bigoplus _{i=1}^s A_n\otimes M_{k_i}$ . So each $A_n$ is unital. In each $M_{k_i}$ , find a projection $e_i$ with rank $l_imn$ , $i=1,2,\dotsc ,s$ . Put $e=\oplus _{i=1}^s 1_{A_n}\otimes e_i$ and $\theta =1-e$ . We will identify $M_{l_imn}$ with $M_{l_im}\otimes M_n$ . Then we have

   1. (i) $\theta x\approx _{\varepsilon } x\theta $ for all $x\in {\cal F}$ ,

   2. (ii) $(1-\theta )x(1-\theta )=exe\in _{\varepsilon } \bigoplus _{i=1}^s A_n\otimes e_i\subset \bigoplus _{i=1}^s \left (A_n\otimes M_{l_im}\right )\otimes 1_n$ and

   3. (iii) $\theta \lesssim a$ , as $d_\tau (1-e)<1/mn<d_\tau (a)$ for all $\tau \in QT(A)$ .

   From this we conclude that A is tracially approximately divisible (see also Proposition 5.3).

2. (2) Note that the Jiang–Su algebra ${\mathscr Z}$ is not approximately divisible, as it has no nontrivial projections. However, by Theorem 5.9, it is tracially approximately divisible.

3. (3) In a subsequent paper [20, Theorem 4.11], at least in the separable case, we show that the converse of Proposition 5.3 also holds. In fact, in [20, Lemma 4.9] we show that a weaker version of Definition 5.2, without mentioning $e_F$ , implies the conditions stated in Proposition 5.3. In other words, in Definition 5.2, any reference to $e_F$ could be omitted. However, the proof is somewhat more involved; we refer the reader to [20] for further discussion.

4. (4) There is also a notion called ‘tracially almost divisibility’ (see [48, Definition 3.5]). That definition uses quasitraces, whereas Definition 5.2 does not mention quasitraces. They are quite different. However, it is not hard to show that tracially approximate divisibility implies tracially almost divisibility. In [20], we show a separable simple $C^*$ -algebra A which is tracially approximately divisible, has strict comparison and stable rank one and has a nice description of its Cuntz semigroup. These imply, in particular, that A has the tracially almost divisible property defined in [48, Definition 3.5]. The converse, in general, does not hold even with strict comparison – for example, $A=C^*_{\mathrm {red}}(F_\infty )$ (see [20, 7.3]).

Proposition 5.3 compare [32, 5.3]

Suppose that A is a simple $C^*$ -algebra which satisfies the following conditions: For any $\varepsilon>0$ , any finite subset ${\cal F}\subset A$ , any $s\in A_+\setminus \{0\}$ and any integer $n\ge 1$ , there are $\theta \in A_+^{\textbf{1}}$ and a $C^*$ -subalgebra $D\otimes M_n\subset A$ such that

1. (i) $\theta x\approx _{\varepsilon } x\theta $ for all $x\in {\cal F}$ ,

2. (ii) ${{(1-\theta )}}x\in _{\varepsilon } D\otimes 1_n$ for all $x\in {\cal F}$ and

3. (iii) $\theta \lesssim s$ .

Then A is tracially approximately divisible.

Proof. Let ${\cal F}\subset A$ a finite subset, $\varepsilon>0$ , $s\in A_+\setminus \{0\}$ and an integer n be given. Suppose that there are a finite subset ${\cal F}'$ and an element $e_F\in A_+^{\textbf{1}}$ such that $e_Fy=y=ye_F$ for all $y\in {\cal F}'$ , and if $x\in {\cal F}$ , there is $y\in {\cal F}'$ such that $\lVert y-x\rVert <\varepsilon /4$ . Without loss of generality, we may assume that ${\cal F}\subset A^{\textbf{1}}$ . We may further assume that ${\cal F}'\subset A^{\textbf{1}}$ .

Let $\delta \in (0,\varepsilon /8)$ be a positive number such that for any elements $z\in A^{\textbf{1}}$ and ${{w}}\in A_+^{\textbf{1}}$ , $\lVert zw-wz\rVert <\delta $ implies that

(e5.1) $$ \begin{align} {{\left\lVert(1-w)^{1/2}z-z(1-w)^{1/2}\right\rVert<\varepsilon/8.}} \end{align} $$

Put ${\cal F}_1={\cal F}\cup \{e_F\}\cup {\cal F}'$ . Suppose that there are $\theta \in A_+^{\textbf{1}}$ and D as in the statement of the proposition, such that (i), (ii) and (iii) hold for $\delta $ (in place of $\varepsilon $ ) and ${\cal F}_1$ (in place of ${\cal F}$ ).

Then in Definition 5.2(3) holds.

Define ${\beta }: A\to A$ by ${\beta }(a):= (1-\theta )^{1/2}a(1-\theta )^{1/2}$ for all $a\in A$ . It is a c.p.c. map. For each $x\in {\cal F}_1$ , define $x_1:=\theta ^{1/2} x\theta ^{1/2} \,\in \mathrm {Her}(\theta )$ . Then $\lVert x_1\rVert \le \lVert x\rVert $ . Note that by the choice of $\delta $ , for all $x\in {\cal F}\cup {\cal F}'$ ,

(e5.2) $$ \begin{align} e_F{\beta}(x)=e_F(1-\theta)^{1/2}x(1-\theta)^{1/2}\approx_{\varepsilon/8}(1-\theta)^{1/2}e_Fx(1-\theta)^{1/2}\approx_{\varepsilon/8} {\beta}(x) \approx_{\varepsilon/4} {\beta}(x)e_F. \end{align} $$

Moreover, for all $x\in {\cal F}_1$ ,

(e5.3) $$ \begin{align} {\beta}(x)=(1-\theta)^{1/2}x(1-\theta)^{1/2} {{ \approx_{\varepsilon/8} (1-\theta)}}x\in_{\delta} D\otimes 1_n. \end{align} $$

So Definition 5.2(2) holds. Also by the choice of $\delta $ , for all $x\in {\cal F}_1$ ,

(e5.4) $$ \begin{align} x=\theta x+(1-\theta)x\approx_{\varepsilon/4} \theta^{1/2}x\theta^{1/2}+(1-\theta)^{1/2}x(1-\theta)^{1/2}=x_1+{\beta}(x). \end{align} $$

Hence Definition 5.2(1) holds. Thus A is tracially approximately divisible.

The following lemma is convenient folklore:

Lemma 5.4. Let $\delta>0$ . There is an integer $N(\delta )\ge 1$ such that for any $C^*$ -algebra A, any $e\in A_+^{\textbf{1}}$ and any $x\in A$ , if $x^*x\le e$ and $xx^*\le e$ , then

(e5.5) $$ \begin{align} e^{1/n}x\approx_{\delta} x\approx_{\delta} xe^{1/n}\mathrm{\ for\ all\ } n\ge N(\delta). \end{align} $$

Proof. Let $\delta>0$ be given. Choose ${{N(\delta )}}\ge 1$ such that

(e5.6) $$ \begin{align} \max\left\{\left\lvert{{\left(1-t^{1/n}\right)^2t}}\right\rvert: t\in [0,1]\right\}<\delta^2 \mathrm{\ for\ all\ } n\ge {{N(\delta)}}. \end{align} $$

Then for any $C^*$ -algebra A, any $e\in A_+^{\textbf{1}}$ and any $x\in A$ satisfying $x^*x\le e$ and $xx^*\le e$ ,

(e5.7) $$ \begin{align} \left\lVert\left(1-e^{1/n}\right)x\right\rVert = \left\lVert\left(1-e^{1/n}\right)xx^*\left(1-e^{1/n}\right)\right\rVert^{1/2} \leq \left\lVert\left(1-e^{1/n}\right)e\left(1-e^{1/n}\right)\right\rVert^{1/2} <\delta \end{align} $$

for all $n\geq N(\delta )$ . Similarly, we also have $\left \lVert x\left (1-e^{1/n}\right )\right \rVert <\delta $ for all $n\geq N(\delta )$ . The lemma follows.

Theorem 5.5. If A is a simple $C^*$ -algebra which is tracially approximately divisible, then every hereditary $C^*$ -subalgebra of A is also tracially approximately divisible.

Proof. Let B be a hereditary $C^*$ -subalgebra of A, ${\cal F}\subset B^{\textbf{1}}$ be a finite subset, $\varepsilon>0$ , $s\in B_+\setminus \{0\}$ be a positive element and $n\ge 1$ be an integer. Suppose also that there exists a finite subset ${\cal F}'\subset B^{\textbf{1}}$ such that $y\in _{\varepsilon /4}{\cal F}'$ for all $y\in {\cal F}$ , and there exists an element $e_F\in B_+^{\textbf{1}}$ such that $e_Fx=x=xe_F$ for all $x\in {\cal F}'$ . Let $g_0, g_1\in C_0([0,1])$ be such that $0\le g_0, g_1\le 1$ , $g_0(0)=0$ , $g_0(t)=1$ for $t\in [1-\varepsilon /64,1]$ and $g_0$ is linear on $[0, 1-\varepsilon /64];$ and $g_1(t)=0$ if $t\in [0,1-\varepsilon /64]$ , $g_1(1)=1$ and $g_1$ is linear on $[1-\varepsilon /64, 1]$ . Put $b_0:=g_0(e_F)$ and $b_1:=g_1(e_F)$ . Then

(e5.8) $$ \begin{align} b_0b_1=b_1=b_1b_0,\qquad b_0\ge e_F, \qquad \lVert b_0-e_F\rVert<\varepsilon/64. \end{align} $$

Since for all $x\in {\cal F'}$ we have $e_Fxx^*=xx^*=xx^*e_F$ and $e_Fx^*x=x^*x=x^*xe_F$ , by the spectral theory, we have $b_ixx^*=xx^*=xx^*b_i$ and $b_ix^*x=x^*x=x^*xb_i$ , $i=0,1$ . It follows that

(e5.9) $$ \begin{align} b_ix=x=xb_i\mathrm{\ and\ } b_ix^*=x^*=x^*b_i\mathrm{\ for\ all\ } x\in {\cal F}', \quad i=0,1. \end{align} $$

Let ${\cal F}_1= \{b_1\}\cup {\cal F}'$ . Choose $\delta>0$ in [15, Lemma 3.3] associated with $\varepsilon /64$ (in place of $\varepsilon $ ) and $\sigma =\varepsilon /64$ . Set $\eta =\min \{\delta /4, \varepsilon /256\}$ .

We choose $N:= N(\eta )\ge 1$ as in Lemma 5.4.

Let $0<\delta _1<\eta /2$ . Moreover, we choose $\delta _1$ sufficiently small that if $C_1\subset C_2$ is any pair of $C^*$ -algebras and $c\in C_2$ with $0\le c\le 1$ and $c\in _{\delta _1} C_1$ , and if $0\le c_1, c_2\le 1$ and $c_1c_2\approx _{\delta _1} c_2\approx _{\delta _1}c_2c_1$ , then

(e5.10) $$ \begin{align} c^{1/N}\in_{\eta} (C_1)_+^{\textbf{1}}\qquad \text{and} \qquad c_1c_2^{1/N}c_1\approx_\eta c_2^{1/N}. \end{align} $$

Since A is tracially approximately divisible, there are $\theta _a\in A_+^{\textbf{1}}$ , a $C^*$ -subalgebra $D_a\otimes M_n\subset A$ and a c.p.c. map ${\beta }: A\to A$ such that

1. (1) $x\approx _{\delta _1/2} x_1+ {\beta }(x) $ such that $\lVert x_1\rVert \le 1$ and $x_1\in \mathrm {Her}(\theta _a)$ for all $x\in {\cal F}_1$ ,

2. (2) ${\beta }(x)\in _{\delta _1/2} D_a\otimes 1_n$ and $ b_0{\beta }(x)\approx _{\delta _1/2} {\beta }(x)\approx _{\delta _1/2} {\beta }(x)b_0 \text { for all } x\in {\cal F}_1$ and

3. (3) $\theta _a\lesssim s$ .

Choose $d(x)\in (D_a\otimes 1_n)^{\textbf{1}}$ such that

(e5.11) $$ \begin{align} \lVert{\beta}(x)-d(x)\rVert<\delta_1\mathrm{\ for\ all\ } x\in {\cal F}_1. \end{align} $$

Let $b_2={\beta }(b_1)^{1/N}$ . By equation (e5.9), ${\beta }(b_1)\ge {{{\beta }(x)^*{\beta }(x)}}$ and ${\beta }(b_1)\ge {\beta }(x){\beta }(x)^*$ for all $x\in {\cal F}'$ (see, for example, [5, Corollary 4.1.3]). By condition (2) here, the choice of N and application of Lemma 5.4,

(e5.12) $$ \begin{align} b_2{\beta}(x)={\beta}(b_1)^{1/N}{\beta}(x) \approx_\eta {\beta}(x)\qquad \mathrm{\ for\ all\ } x\in {\cal F}'. \end{align} $$

Recall that ${\beta }(b_1)\in _{\delta _1} D_a\otimes 1_n$ . By the choice of $\delta _1$ , we may choose $d\in (D_a\otimes 1_n)_+$ such that

(e5.13) $$ \begin{align} \lVert d-b_2\rVert<\eta. \end{align} $$

Then, with $b:=b_0b_2b_0$ , by the second part of formula (e5.10),

(e5.14) $$ \begin{align} \lVert d-b\rVert<2\eta \qquad\text{and}\qquad f_{\varepsilon/64}(d)d\approx_{\varepsilon/64} d\approx_{2\eta} b. \end{align} $$

By the choice of $\eta $ , applying [15, Lemma 3.3] yields an isomorphism

$$ \begin{align*} \varphi: \overline{f_{\varepsilon/64}(d)(D_a\otimes M_n)f_{\varepsilon/64}(d)}\to \overline{bAb} \subset B \end{align*} $$

such that

(e5.15) $$ \begin{align} \lVert \varphi(y)-y\rVert<\varepsilon/64\lVert y\rVert\mathrm{\ for\ all\ } y\in \overline{f_{\varepsilon/64}(d)(D\otimes 1_n)f_{\varepsilon/64}(d)}. \end{align} $$

Note that $\overline {f_{\varepsilon /64}(d)(D_a\otimes M_n)f_{\varepsilon /64}(d)}\cong D_1\otimes M_n$ and $\overline {f_{\varepsilon /64}(d)(D_a\otimes 1_n)f_{\varepsilon /64}(d)}\cong D_1\otimes 1_n$ for some $C^*$ -subalgebra $D_1\subset D_a$ . Let $D_b=\varphi (D_1)$ . Define a c.p.c. map ${\alpha }: B\to {{B}}$ by

(e5.16) $$ \begin{align} {\alpha}(y):= b{\beta}(y)b\mathrm{\ for\ all\ } y\in {{B}}. \end{align} $$

Then, for all $x\in {\cal F}_1$ , by formulas (e5.14) and (e5.11),

(e5.17) $$ \begin{align} {\alpha}(x)&=b{\beta}(x)b\approx_{2\left(2\eta+\varepsilon/64\right)} f_{\varepsilon/64}(d) d{\beta}(x)d f_{\varepsilon/64}(d) \hspace{2pc} \end{align} $$

(e5.18) $$ \begin{align} \hspace{3pc} &\approx_{\delta_1} f_{\varepsilon/64}(d)d d(x) df_{\varepsilon/64}(d)\in_{\varepsilon/64} D_b\otimes 1_n\subset \overline{bAb}\subset B. \end{align} $$

If $y\in {\cal F}$ , choose $x\in {\cal F}'$ such that $\lVert y-x\rVert <\varepsilon /4$ . Then ${\alpha }(y)\approx _{\varepsilon /4} {\alpha }(x)\in _{\varepsilon /4} D_b\otimes 1_n$ . Define $y_1=b_0x_1b_0$ . Then, by conditions (1) and (2) and equation (e5.12),

(e5.19) $$ \begin{align} y&\approx_{\delta_1/2+\varepsilon/4} b_0(x_1+{\beta}(x))b_0=y_1+b_0{\beta}(x)b_0 \end{align} $$

(e5.20) $$ \begin{align} &\approx_{2\eta} y_1+b_0b_2{\beta}(x)b_2b_0 \hspace{6pc} \end{align} $$

(e5.21) $$ \begin{align} &\approx_{2\delta_1}y_1+b_0b_2b_0{\beta}(x)b_0b_2b_0=y_1+{\alpha}(x) \end{align} $$

(e5.22) $$ \begin{align} &\approx_{\varepsilon/4} y_1+{\alpha}(y) \quad \text{for all }y\in {\cal F}. \hspace{3.5pc} \end{align} $$

Note that $\delta _1/2+\varepsilon /4+2\eta +2\delta _1+\varepsilon /4<\varepsilon $ . Put $\delta _2:=3\delta _1/2+2\eta $ . Then $0<\delta _2<5\varepsilon /256$ . Also for all $y\in {\cal F}$ ,

(e5.23) $$ \begin{align} e_F{\alpha}(y)\approx_{\varepsilon/4}e_F{\alpha}(x)&=e_F b_0b_2b_0{\beta}(x)b_0b_2b_0\approx_{\delta_2} e_F{\beta}(x)\approx_{\varepsilon/64} b_0{\beta}(x)\approx_{\delta_1} {\beta}(x) \end{align} $$

(e5.24) $$ \begin{align} &\approx_{\delta_1} {\beta}(x)b_0\approx_{\varepsilon/64}{\beta}(x)e_F\approx_{\delta_2+\varepsilon/4}{\alpha}(y)e_F \end{align} $$

(recall $\lVert b_0-e_F\rVert <\varepsilon /64$ ). Put $\theta _b=b_0\theta _a b_0$ . Then $y_1\in \overline {\theta _bB\theta _b}$ . Moreover,

(e5.25) $$ \begin{align} \theta_b\lesssim \theta_a\lesssim s. \end{align} $$

From formulas (e5.22), (e5.18), (e5.24) and (e5.25), the theorem follows.

Lemma 5.6. Let A be a $C^*$ -algebra and set $n\in \mathbb {N}$ . Let $e_1,\dotsc ,e_n\in A_+$ be mutually orthogonal nonzero positive elements. Assume $d_1,\dotsc , d_n\in A_+$ such that $d_i\lesssim e_i$ ( $i=1,\dotsc ,n$ ), and $e_i d_j=0$ whenever $i\le j$ and $i,j=1,\dotsc ,n$ . Then for any $a\in {{\overline {d_1Ad_1+\dotsb + d_nAd_n}}}$ and any $\varepsilon>0$ , there are nilpotent elements $x,y\in A$ such that $\lVert a-yx\rVert <\varepsilon $ .

Proof. Set $a\in {{\overline {d_1Ad_1+\dotsb +d_nAd_n}}}$ and fix $\varepsilon>0$ . Then there exist $a_1,\dotsc ,a_n\in A$ and $\delta>0$ such that $a\approx _{\varepsilon } f_\delta (d_1)a_1f_\delta (d_1)+\dotsb +f_{\delta }(d_n)a_nf_\delta (d_n)$ . Set $x_1,\dotsc ,x_n\in A$ such that $x_i^*x_i=f_\delta (d_i)$ and $x_ix_i^*\in \overline {e_iAe_i}$ , $i=1,\dotsc ,n$ (see [40, Proposition 2.4]). For $i,j\in \{1,\dotsc ,n\}$ and $i\leq j$ , $e_id_j=0$ implies $x_j^*x_jx_ix_i^*=0$ , thus

(e5.26) $$ \begin{align} x_jx_i=0 \quad(i\leq j). \end{align} $$

Claim 1: ${{(x_1+x_2+\dotsb +x_n)^{n+1}=0}}$ .

Proof of Claim 1: Note that $(x_1+x_2+\dotsb +x_n)^{n+1}$ is a sum of $n^{n+1}$ terms with the form $x_{k_1}x_{k_2}\dotsm x_{k_{n+1}} (k_1,\dotsc ,k_{n+1}\in \{1,\dotsc ,n\})$ . Assume $x_{k_1}x_{k_2}\dotsm x_{k_n+1}\neq 0$ ; then $x_{k_i}x_{k_{i+1}}\neq 0$ ( $i=1,\dotsc ,n$ ). By equation (e5.26), it follows that $k_{i+1}\leq k_i-1$ ( $i=1,\dotsc ,n$ ). In particular, $k_{n+1}\le k_n-1$ . Then $k_{n+1}\le k_n-1\le k_{n-1}-2$ . An induction implies that $k_{n+1}\leq k_1-n\leq 0$ , which gives a contradiction. Thus all $n^{n+1}$ terms of the form $x_{k_1}x_{k_2}\dotsm x_{n+1}$ are zero. It follows that $(x_1+x_2+\dotsb +x_n)^{n+1}=0$ .

Claim 2: $\left (f_\delta (d_1)a_1x_1^*+\dotsb +f_{\delta }(d_n)a_nx_n^*\right )^{n+1}=0$ .

Proof of Claim 2: Let $y_i=f_\delta (d_i)a_ix_i^*$ ( $i=1,\dotsc ,n$ ). For $i\leq j$ , using equation (e5.26), we have

(e5.27) $$ \begin{align} y_iy_j = f_\delta(d_i)a_ix_i^*f_\delta\left(d_j\right)a_jx_j^* = f_\delta(d_i)a_ix_i^*\left(x_j^*x_j\right)a_jx_j^*=f_\delta(d_i)a_i\left(x_jx_i\right)^*x_ja_jx_j^* =0. \end{align} $$

Then, as in the proof of Claim 1, we have $(y_1+\dotsb +y_n)^{n+1} =0$ . Claim 2 follows.

Let $x=x_1+\dotsb +x_n$ and let $y=y_1+\dotsb +y_n=f_\delta (d_1)a_1x_1^*+\dotsb +f_{\delta }(d_n)a_nx_n^*$ . Then by Claims 1 and 2, both x and y are nilpotent elements. For $i,j\in \{1,\dotsc ,n\}$ and $i\neq j$ , $e_ie_j=0$ implies $x_ix_i^*x_jx_j^*=0$ , thus $ x_i^*x_j=0. $ Then $yx=f_\delta (d_1)a_1f_\delta (d_1)+\dotsb +f_{\delta }(d_n)a_nf_\delta (d_n)\approx _\varepsilon a$ .

Recall that a non-unital $C^*$ -algebra is said to almost have stable rank one if for every hereditary $C^*$ -subalgebra $B\subset A$ , B lies in the closure of invertible elements of $\widetilde B$ [38, Definition 3.1].

Theorem 5.7. Let A be a simple $C^*$ -algebra which is tracially approximately divisible. Suppose that A is stably finite and $W(A)$ is almost unperforated. Then A has stable rank one if it is unital, or almost has stable rank one if it is not unital.

Proof. We assume that A is infinite-dimensional. Fix an element $x\in A$ and fix $\varepsilon>0$ . We may assume that x is not invertible. Since A is finite, x is not one-sided invertible. To show that x is a norm limit of invertible elements, it suffices to show that $ux$ is a norm limit of invertible elements for some unitary $u\in {\widetilde A}$ . Note that since A is simple, ${\widetilde A}$ is prime. Thus, by [39, Proposition 3.2, Lemma 3.5], we may assume that there is $a'\in \widetilde A_+\backslash \{0\}$ and $a'x=xa'=0$ . There is $e\in A_+$ such that $a'ea'\not =0$ . Put $a=a'ea'$ .

Let $B_0=\{z\in A: az=za=0\}$ . Then $x\in B_0$ , and $B_0$ is a hereditary $C^*$ -subalgebra of A. There is $e_b'\in {B_0}_+$ with $\lVert e_b\rVert =1$ such that $e_b'xe_b'\approx _{\varepsilon /64}x$ . So $f_{\varepsilon /64}\left (e_b'\right )xf_{\varepsilon /64}\left (e_b'\right )\approx _{\varepsilon /16}x$ . Put $e_b=f_{\varepsilon /64}\left (e_b'\right )$ and $B=\mathrm {Her}(e_b)$ . Without loss of generality, we may further assume that $x\in B$ .

Since we assume that A is infinite-dimensional, $\overline {aAa}$ contains nonzero positive elements $a_0,a_1$ such that $a_0a_1=0$ .

Since A is simple, there is $c\in A$ such that $e_bc(a_1)^{1/2}\not =0$ (see the proof of [12, 1.8]).

Note that since $e_b\in \mathrm {Ped}(B)$ , we have $\mathrm {Ped}(B)=B$ (see, for example, [3, II.5.4.2]). It follows that there are $y_1,y_2,\dotsc ,y_m\in B$ such that

(e5.28) $$ \begin{align} \sum_{i=1}^m y_i^*e_bca_1c^*e_by_i=e_b. \end{align} $$

It then follows that $\langle e_b\rangle \le m\langle a_1\rangle $ . Put $n=2m$ .

For any $z_1,z_2,\dotsc ,z_n\in B_+$ which are n mutually orthogonal and mutually equivalent positive elements,

$$ \begin{align*}n\langle z_1\rangle \le \langle e_b\rangle \le m\langle a_1\rangle. \end{align*} $$

Since $W(A)$ is almost unperforated,

(e5.29) $$ \begin{align} z_1\lesssim a_1. \end{align} $$

Since B is a hereditary $C^*$ -subalgebra of A, by Theorem 5.5, B is also tracially approximately divisible. There are $b\in B_+^{\textbf{1}}$ , a $C^*$ -subalgebra $D\otimes M_n\subset B$ and a c.p.c. map ${\beta }: A\to A$ such that

1. (1) $x\approx _{\varepsilon /8} x_0+{\beta }(x)$ , where $x_0\in \overline {bAb}$ ,

2. (2) ${\beta }(x)\in _{\varepsilon /8} D\otimes 1_n$ and

3. (3) $b\lesssim a_0$ .

Thus, there is $x_1\in D\setminus \{0\}$ such that

(e5.30) $$ \begin{align} \lVert x-(x_0+ x_1\otimes 1_n)\rVert<\varepsilon/4. \end{align} $$

Choose a positive element $d\in D$ such that

(e5.31) $$ \begin{align} \lVert dx_1d-x_1\rVert<\varepsilon/4. \end{align} $$

By the choice of n, we have $d\otimes e_{1,1}\lesssim a_1$ , where $\left \{e_{i,j}\right \}$ forms a system of matrix units for $M_n$ .

Define $g_1:= a_0$ , $g_2 := a_1$ , $g_{2+i}:= d\otimes e_{i,i}$ ( $i=1,\dotsc ,n-1$ ).

Define $h_1:= b$ , $h_{1+i}:= d\otimes e_{i,i}$ ( $i=1,\dotsc ,n$ ).

Note that $h_i\lesssim g_i$ ( $i=1,\dotsc ,n+1$ ) and $g_ih_j=0$ , if $i\le j$ and $i,j=1,\dotsc ,n+1$ . Note that $x_0+dx_1d\otimes 1_n\in \overline {h_1Ah_1}+\overline {h_2Ah_2}+\dotsb +\overline {h_{n+1}Ah_{n+1}}$ . Then by Lemma 5.6, there are nilpotent elements $v,w\in A$ such that $x_0+dx_1d\otimes 1_n\approx _{\varepsilon /4}vw$ . Choose $\delta>0$ such that $vw\approx _{\varepsilon /4}(v+\delta )(w+\delta )$ . Since $v,w$ are nilpotent elements, $v+\delta $ and $w+\delta $ are invertible. Then, combining formulas (e5.30) and (e5.31),

(e5.32) $$ \begin{align} x\approx_{\varepsilon/4} x_0+x_1\otimes 1_n \approx_{\varepsilon/4} x_0+dx_1d\otimes 1_n\approx_{\varepsilon/2} (v+\delta)(w+\delta) \in GL\left(\widetilde A\right). \end{align} $$

Therefore we have shown that $x\in \overline {GL\left (\widetilde A\right )}$ . Thus, in the case that A is unital, A has stable rank one. Since, by Theorem 5.5, this works for every hereditary $C^*$ -subalgebra of A, A almost has stable rank one in the case that A is not unital.

Remark 5.8. Under the assumption of Theorem 5.7, if $x\in A$ is not invertible, then there is a unitary $u\in \tilde A$ such that $(ux)e=e(ux)=0$ for some $e\in A_+\setminus \{0\}$ . The proof shows that $ux$ can be approximated by products of two nilpotents in A. The idea of the proof is taken from the proof of [15, Lemma 11.1], which originates from that of [38, Lemma 2.1] and [39].

In a subsequent paper [20], we will show that a separable simple $C^*$ -algebra which is tracially approximately divisible has strict comparison for positive elements. So there is a redundancy in the assumption of Theorem 5.7.

Theorem 5.9. Let A be a simple $C^*$ -algebra. If A is essentially tracially in ${\cal C}_{\mathscr Z}$ , then it is tracially approximately divisible.

Proof. We assume that A is infinite-dimensional. Let A be a simple $C^*$ -algebra which is e. tracially in ${\cal C}_{\mathscr Z}$ . By [42, Theorem 4.5], every ${\mathscr Z}$ -stable $C^*$ -algebra B has almost unperforated $W(B)$ (see Remark 2.5). Then, by Theorem 4.3, $W(A)$ is almost unperforated. Let $\varepsilon>0$ , ${\cal F}\subset A^{\textbf{1}}$ a finite subset, $a\in A_+\setminus \{0\}$ and $n\ge 1$ an integer be given. Since A is infinite dimensional, choose $a_1, a_2\in \mathrm {Her}(a)_+\setminus \{0\}$ such that $a_1a_2=a_2a_1=0$ .

There are $e_A\in A_+^{\textbf{1}}$ and $\delta>0$ such that

(e5.33) $$ \begin{align} f_\delta(e_A)x\approx_{\varepsilon/4} x\approx_{\varepsilon/4} xf_\delta(e_A)\mathrm{\ for\ all\ } x\in {\cal F}. \end{align} $$

Note that by Theorem 5.5, $A_1:=\overline {f_{\delta /2}(e_A)Af_{\delta /2}(e_A)}$ is also a ( $\sigma $ -unital) simple $C^*$ -algebra which is e. tracially in ${\cal C}_{\mathscr Z}$ (as ${\cal C}_{\mathscr Z}$ has property (H); see [46, Corollary 3.1]).

Note also that $f_{\delta /2}(e_A)af_{\delta /2}(e_A)\lesssim a$ . To simplify notation, by replacing x by $f_\delta (e_A)xf_\delta (e_A)$ for all $x\in {\cal F}$ , a by $f_{\delta /2}(e_A)af_{\delta /2}(e_A)$ and $a_i$ by $f_{\delta /2}(e_A)a_if_{\delta /2}(e_A)$ ( $i=1,2$ ), without loss of generality we may assume that $x, a, a_1, a_2\in A_1$ . We may also assume, without loss of generality,

(e5.34) $$ \begin{align} e_1x=x=xe_1\mathrm{\ for\ all\ } x\in {\cal F} \end{align} $$

for some strictly positive element $e_1\in A_1^{1} $ . Note that $f_{\delta /2}(e_A)\in \mathrm {Ped}(A)$ . Therefore $A_1$ is algebraically simple and $f_{\delta /2}(e_A)$ is a strictly positive element of $A_1$ . There are an integer $l\ge 1$ and $x_i\in A_1$ , $i=1,2,\dotsc ,l$ , such that

(e5.35) $$ \begin{align} \sum_{i=1}^l x_i^*a_1x_i=e_1. \end{align} $$

Set ${\cal F}_1={\cal F}\cup \{e_1\}$ . Choose $0<\eta <\varepsilon /2$ such that if $\theta '\in A_+^{\textbf{1}}$ with $\lVert \theta 'x-x\theta '\rVert <\eta $ , then

(e5.36) $$ \begin{align} (\theta')^{1/2}x\approx_{\varepsilon/2} x(\theta')^{1/2}\mathrm{\ for\ all\ } x\in {\cal F}_1. \end{align} $$

There exist $\theta _1\in A_+^{\textbf{1}}$ and a ${\mathscr Z}$ -stable $C^*$ -subalgebra B of $A_1$ such that

1. (i) $\lVert \theta _1x-x\theta _1\rVert <\eta /64$ and $\left \lVert (1-\theta _1)^{1/2}x-x(1-\theta _1)^{1/2}\right \rVert <\eta /64$ for all $x\in {\cal F}_1$ ,

2. (ii) $(1-\theta _1)^{1/2}x(1-\theta _1)^{1/2}, (1-\theta _1)^{1/2}x, x(1-\theta _1)^{1/2}, (1-\theta _1)x, x(1-\theta _1), (1-\theta _1)x(1-\theta _1)\in _{\eta /64} B$ for all $x\in {\cal F}_1$ and

3. (iii) $\theta _1\lesssim a_2$ .

Let

$$ \begin{align*} {\cal F}_2&=\left\{(1-\theta_1)^{1/2}x(1-\theta_1)^{1/2}, (1-\theta_1)^{1/2}x, x(1-\theta_1)^{1/2}, (1-\theta)x, x(1-\theta_1), (1-\theta_1)x(1-\theta_1): \right. \nonumber \\ &\left. \quad x\in {\cal F}\right\}. \end{align*} $$

For each $f\in {\cal F}_2$ , fix $b(f)\in B$ such that $\lVert b(f)\rVert \le 1$ and

(e5.37) $$ \begin{align} \lVert f-b(f)\rVert<\eta/32. \end{align} $$

Let ${\cal G}=\{b(f): f\in {\cal F}_2\}$ . We write $B=C\otimes {\mathscr Z}$ . Since ${\mathscr Z}$ is strongly self-absorbing, without loss of generality we may assume that there is a finite subset ${\cal G}_1\subset C$ such that ${\cal G}=\{y\otimes 1_{\mathscr Z}: y\in {\cal G}_1\}\subset C\otimes 1_{\mathscr Z}$ . To further simplify notation, without loss of generality we may assume that there exists a strictly positive element $e_C\in C$ such that

(e5.38) $$ \begin{align} e_by=y=ye_b\mathrm{\ for\ all\ } y\in {\cal G}_1, \end{align} $$

where $e_b=e_C\otimes 1_{\mathscr Z}$ .

For any integer n, choose m such that $m> l$ and n divides m. Let $\psi : M_m\to {\mathscr Z}$ be an order $0$ c.p.c. map such that

(e5.39) $$ \begin{align} 1_{\mathscr Z}-\psi(1_m)\lesssim_{\mathscr Z} \psi\left(e_{1,1}\right) \end{align} $$

(see [43, Proposition 5.1(iv) implying (ii)]). Define $\varphi : M_m\to B$ by $\varphi (c):= e_C\otimes \psi (c)$ for all $c\in M_m$ . Set

(e5.40) $$ \begin{align} \theta_2:=e_b-\varphi(1_m)=e_C\otimes 1_{\mathscr Z}-e_C\otimes \psi(1_m)=e_C\otimes (1_{\mathscr Z}-\psi(1_{{m}}))\lesssim_{B} e_C\otimes \psi\left(e_{1,1}\right). \end{align} $$

Note that

(e5.41) $$ \begin{align} \theta_2 g=g\theta_2\mathrm{\ for\ all\ } g\in {\cal G}. \end{align} $$

It follows from formulas (e5.37) and (e5.41) that for any $y\in {\cal F}_2$ ,

(e5.42) $$ \begin{align} \theta_2y\approx_{\varepsilon/32} y\theta_2. \end{align} $$

Define $D:= \overline {e_Cce:C\otimes \psi \left (e_{1,1}\right )}$ and $D'$ the $C^*$ -subalgebra generated by

(e5.43) $$ \begin{align} \{e_Cce:C\otimes \psi(z): c\in C\mathrm{\ and\ } z\in M_m\}. \end{align} $$

Recall that $\psi $ gives a homomorphism $H: C^*(\psi (1_m))\otimes M_m \to {\mathscr Z}$ such that $H(\iota \otimes g)=\psi (g)$ for all $g\in M_m $ , where $\iota (t)=t$ for $t\in \mathrm {sp}(\psi (1_m))$ (see [49, Corollary 4.1]). It follows that $D'\cong D\otimes M_m$ . Define ${\beta }_1: A\to A$ by

(e5.44) $$ \begin{align} {\beta}_1(y):= \left(1_{\widetilde{A}}-\theta_2\right)^{1/2}y\left(1_{\widetilde{A}}-\theta_2\right)^{1/2}\mathrm{\ for\ all\ } y\in A \end{align} $$

(where $1_{\widetilde {A}}$ denotes the identity of $\widetilde {A}$ when A is not unital and is the identity of A if A has one). Note also that $(1-\theta )^{1/2}$ is an element which has the form $1+f_1(\theta )$ for $f_1(t)=(1-t)^{1/2}-1\in C_0([0,1])^{\textbf{1}}$ . If $g=y\otimes 1_{\mathscr Z}\in {\cal G}$ , then (noting that $y\in {\cal G}_1\subset C$ , and seeing equation (e5.38)),

(e5.45) $$ \begin{align} {\beta}_1(g)&=(1-\theta_2)g =g -e_C\otimes (1_{\mathscr Z}-\psi(1_m))g \hspace{9pc} \end{align} $$

(e5.46) $$ \begin{align} &=(e_C\otimes 1_{\mathscr Z})g-e_C\otimes(1_{\mathscr Z}- \psi(1_m))g \hspace{7.6pc} \end{align} $$

(e5.47) $$ \begin{align} &=(e_C\otimes \psi(1_m)) (y\otimes 1_{\mathscr Z})=\left(e_C^{1/2}ye_C^{1/2}\right)\otimes \psi(1_m)\in D\otimes 1_m. \end{align} $$

Define a c.p.c. map ${\beta }: A\to A$ by

(e5.48) $$ \begin{align} {\beta}(x):= {\beta}_1\left((1-\theta_1)^{1/2}x (1-\theta_1)^{1/2}\right)\mathrm{\ for\ all\ } x\in A. \end{align} $$

For $x\in {\cal F}$ , let $f=(1-\theta _1)^{1/2}x(1-\theta _1)^{1/2}$ . Then, by formula (e5.37),

(e5.49) $$ \begin{align} {\beta}(x)={\beta}_1\left((1-\theta_1)^{1/2}x (1-\theta_1)^{1/2}\right)\approx_{\eta/32} {\beta}_1(b(f))\in D\otimes 1_m. \end{align} $$

Put $\theta =\theta _1+(1-\theta _1)^{1/2}\theta _2(1-\theta _1)^{1/2}$ . We have

(e5.50) $$ \begin{align} 0\le \theta\le \theta_1+(1-\theta_1)^{1/2}(1-\theta_1)^{1/2}=1. \end{align} $$

For $x\in {\cal F}$ , let $f'=(1-\theta _1)x$ . Recall that we assume that $b(f')=y'\otimes 1_{\mathscr Z}$ for some $y'\in C^{\textbf{1}}$ . Then for $x\in {\cal F}$ , applying formulas (e5.37) and (e5.41) repeatedly, we have

(e5.51) $$ \begin{align} (1-\theta)x&=(1-\theta_1)x- (1-\theta_1)^{1/2}\theta_2(1-\theta_1)^{1/2} x \hspace{6.5pc} \end{align} $$

(e5.52) $$ \begin{align} &\approx_{\eta/32} (1-\theta_1)x- (1-\theta_1)x\theta_2=(1-\theta_1)x(1-\theta_2) \end{align} $$

(e5.53) $$ \begin{align} &\approx_{\eta/32} b(f')(1-\theta_2)=(1-\theta_2)^{1/2} b(f')(1-\theta_2)^{1/2} \end{align} $$

(e5.54) $$ \begin{align} \hspace{2pc} &={\beta}_1(b(f'))\approx_{\eta/32} {\beta}_1\left((1-\theta_1)^{1/2}x(1-\theta_1)^{1/2}\right)={\beta}(x). \end{align} $$

From equations (e5.49) and (e5.54), we have

(e5.55) $$ \begin{align} (1-\theta)x \in_{\eta/8} D\otimes 1_m\mathrm{\ for\ all\ } x\in {\cal F}. \end{align} $$

Recall that $(1-\theta _1)^{1/2}x, x(1-\theta _1)^{1/2}, (1-\theta _1)^{1/2}x(1-\theta _1)^{1/2} \in {\cal F}_2$ . Hence, for $x\in {\cal F}$ , by (i) above and formula (e5.42),

(e5.56) $$ \begin{align} \hspace{-0.5pc} \theta x &=\left(\theta_1+(1-\theta_1)^{1/2}\theta_2(1-\theta_1)^{1/2}\right) x\approx_{2\eta/64} x\theta_1+(1-\theta_1)^{1/2} \theta_2 x(1- \theta_1)^{1/2} \end{align} $$

(e5.57) $$ \begin{align} & \hspace{1pc} \approx_{\eta/32} x\theta_1+(1-\theta_1)^{1/2}x(1-\theta_1)^{1/2}\theta_2 \approx_{\eta/32} x \theta_1+\theta_2(1-\theta_1)^{1/2}x(1-\theta_1)^{1/2} \end{align} $$

(e5.58) $$ \begin{align} &\approx_{\eta/32} x\theta_1+(1-\theta_1)^{1/2} x\theta_2(1-\theta_1)^{1/2} \hspace{12pc} \end{align} $$

(e5.59) $$ \begin{align} &\approx_{\eta/64} x\theta_1+ x(1-\theta_1)^{1/2}\theta_2(1-\theta_1)^{1/2}=\theta x. \hspace{10pc} \end{align} $$

Note that by formulas (e5.40) and (e5.35), in $W(A)$ we have

(e5.60) $$ \begin{align} m\langle \theta_2\rangle &=m\langle e_C\otimes (1_{\mathscr Z}- \psi(1_m))\rangle \hspace{11.5pc} \end{align} $$

(e5.61) $$ \begin{align} & \hspace{2pc} \le m\left\langle e_C\otimes \psi\left(e_{1,1}\right)\right\rangle \le \langle e_C\otimes \psi(1_m)\rangle\le \langle e_C\otimes 1_{\mathscr Z}\rangle\le l \langle a_1 \rangle. \end{align} $$

Therefore (recall that $W(A)$ is almost unperforated), since $l<m$ ,

(e5.62) $$ \begin{align} \theta_2\lesssim a_1. \end{align} $$

It follows (noting that $a_1a_2=a_2a_1=0$ ) that

(e5.63) $$ \begin{align} \theta=\theta_1+(1-\theta_1)^{1/2}\theta_2(1-\theta_1)^{1/2}\lesssim a_2+a_1\lesssim a. \end{align} $$

Finally, the theorem follows from formulas (e5.59), (e5.55) and (e5.63), the fact that $D\otimes 1_n$ is embedded into $D\otimes 1_m$ unitally (as n divides m) and Proposition 5.3.

Corollary 5.10. Let A be a simple $C^*$ -algebra which is e. tracially in ${\cal C}_{\mathscr Z}$ . If A is not purely infinite, then it has stable rank one if it is unital and almost has stable rank one if it is not unital.

Proof. By Theorem 5.9, A is tracially approximately divisible. By Theorem 5.1, if A is not purely infinite, then it has strict comparison for positive elements. It follows then from Theorem 5.7 that A has stable rank one if it is unital and almost has stable rank one if it is not unital.

Remark 5.11. For the rest of this paper, we will present nonamenable examples of $C^*$ -algebras which are possibly stably projectionless and are essentially tracially in the class ${\cal C}_{\mathscr Z}$ , the class of ${\mathscr Z}$ -stable $C^*$ -algebras.

6. Construction of $A_z^C$

In this section we first fix a separable residually finite-dimensional (RFD) $C^*$ -algebra C, which may not be exact.

Let B be the unitisation of $C_0((0,1], C)$ . Since $C_0((0,1], C)$ is contractible, $V(B)=\mathbb {N}\cup \{0\}$ , $K_0(B)=\mathbb {Z}$ and $K_1(B)=\{0\}$ .

Let us make the convention that B includes the case that $C=\{0\}$ – that is, $B=\mathbb {C}$ .

Let $\mathfrak {p}=p_1^{r_1}\cdot p_2^{r_2}\cdot \dotsb $ be a supernatural number, where $p_1, p_2,\dotsc $ is a sequence (possibly finite) of distinct prime numbers and $r_i\in \mathbb {N}\cup \{\infty \}$ . Denote by $\mathbb D_{\mathfrak {p}}$ the subgroup of $\mathbb {Q}$ generated by finite sums of rational numbers of the form ${\frac {m}{p_j^i}}$ , where $m\in \mathbb {Z}$ and $i\in \mathbb {N}\cap \left [1, r_j\right ]$ .

Denote by $M_{\mathfrak {p}}$ the UHF-algebra associated with the supernatural number $\mathfrak {p}$ .

The following is a result of Dădărlat [13]:

Theorem 6.1. Fix a supernatural number $\mathfrak {d}$ . There is a unital simple $C^*$ -algebra $A_{\mathfrak {d}}$ which is an inductive limit of $M_{m(n)}(B)$ with injective and unital connecting maps such that

$$ \begin{align*} \left(K_0(A_{\mathfrak{d}}), K_0(A_{\mathfrak{d}})_+, \left[1_{{A_{\mathfrak{d}}}} \right]\right)=(\mathbb D_{\mathfrak{d}}, {\mathbb D_{\mathfrak{d}}}_+,1), \end{align*} $$

$K_1(A_{\mathfrak {d}})=\{0\}$ , and $A_{\mathfrak {d}}$ has a unique tracial state and tracial rank zero.

Proof. This is taken from [13]; we retain the notation used there. For the supernatural number $\mathfrak {d}$ , there is a standard Bratteli system $\left \{B, \underline {\pi }\right \}$ given by Glimm. We use Dădărlat’s restricted system as defined in [13, Definition 3]. Let $D=AF\left (\underline {\pi }\right )$ . Then D is the UHF-algebra with $(K_0(D), K_0(D)_+, [1_D])=(\mathbb D_{\mathfrak {d}}, {\mathbb D_{\mathfrak {d}}}_+,1)$ . Set $A_{\mathfrak {d}}:=B\left (\underline {\pi }\right )$ as in [13, Proposition 8]. Note that [13, Definition 3(ii)] implies that the connecting maps in the restricted system are injective (see also [13, proof of Proposition 8]). The proof of [13, Proposition 9] shows that $B\left (\underline {\pi }\right )$ is a unital simple $C^*$ -algebra of real rank zero and stable rank one, $\left (K_0(A_{\mathfrak {d}}), K_0(A_{\mathfrak {d}})_+, \left [1_{{A_{\mathfrak {d}}}} \right ]\right )=(\mathbb D_{\mathfrak {d}}, {\mathbb D_{\mathfrak {d}}}_+,1)$ , and has a unique tracial state. Note also that since $K_1(B)=0$ , $K_1\left (B\left (\underline {\pi }\right )\right )=0$ . So $K_1(A_{\mathfrak {d}})=0$ . The fact that $A_{\mathfrak {d}}$ has tracial rank zero is also known and, for example, follows from [28, Theorem 3.7.9].

We will review the construction of $A_{\mathfrak {d}}$ and introduce some notation for our construction.

Definition 6.2. Fix a supernatural number $\mathfrak {d}$ . Choose a Bratteli system $AF\left (\underline {\pi }\right )$ (see [13, Definition 2]) for $M_{\mathfrak {d}}$ given by Glimm. Recall that $B= C_0((0,1], C)^{\sim }$ . Following Dădărlat’s construction (see [13, Definition 3, proof of Proposition 8]), one may write $A_{\mathfrak {d}}=\lim _{n\to \infty }\left (M_{d_n'}(B), \delta _n'\right )$ , $d_{n+1}'=d_n\cdot d_n'$ , where $d_n, d_{n}'>1$ are integers, $\delta _n': M_{d_n'}(B)\to M_{d_{n+1}'}(B)$ is defined by

(e6.1) $$ \begin{align} \delta_n'(f):= \begin{pmatrix} f & 0\\ 0 & \gamma_n(f)\end{pmatrix} \mathrm{\ for\ all\ } f\in M_{d_n'}(B) \end{align} $$

and $\gamma _n: B\to M_{d_{n}-1}$ is a unital homomorphism, a $d_{n}-1$ -dimensional representation (we then use $\gamma _n$ for the extension $\gamma _n\otimes \operatorname {id}_{d_n'}: M_{d_n'}(B)\to M_{\left (d_{n}-1\right )d_n'}$ ) which also has the form described in the proof of [13, Proposition 8]. By that proof, this can always be done.

In the Bratteli system $AF\left (\underline {\pi }\right )$ , we may also assume, by passing to a subsequence, that

(e6.2) $$ \begin{align} \lim_{n\to\infty}d_{n}=\infty. \end{align} $$

Also, we assume for any n that $\{\gamma _m: m\ge n\}$ is a separating sequence of finite-dimensional representations. For a more specific construction of $A_{\mathfrak {d}}$ , readers are referred to [13, Definition 3, Proposition 8, Section 3].

It is important that for any $\tau \in T(B)$ ,

(e6.3) $$ \begin{align} \lim_{n\to\infty}\left\lvert\tau\circ \delta_n'(a)-\tau(\gamma_n(a))\right\rvert=0\mathrm{\ for\ all\ } a\in M_{d_n'}(B). \end{align} $$

(Note that by $\tau $ we mean $\tau \otimes \mathrm {tr}_{d^{\prime }_{n}}$ , where $\mathrm {tr}_{d_{n}'}$ is the tracial state of $M_{d_{n}'}$ .)

Consider $\delta _{m,n}':={\delta }_{n-1}'\circ \delta _{n-2}'\circ \dotsb \circ \delta _m': M_{d_m'}(B)\to M_{d_n'}(B)$ . Then we may write

(e6.4) $$ \begin{align} \delta_{m,n}'(f)=\begin{pmatrix} f & 0\\ 0 & \gamma_{m,n}(f)\end{pmatrix} \mathrm{\ for\ all\ } f\in M_{d_m'}(B), \end{align} $$

where $\gamma _{m,n}: B\to M_{d_n'/d^{\prime }_m-1} $ is a finite-dimensional representation. (In the rest of the paper, we also use $\gamma _{m,n}:= \gamma _{m,n}\otimes \operatorname {id}_N : M_{N}(B)\to M_{\left (d_n'/d^{\prime }_m -1\right )N}$ for all integers $N\ge 1$ .) Therefore, if we fix a finite subset ${\cal F}_m\subset M_{d_m'}(B)$ , we may assume that for any $a\in {\cal F}_m\setminus \{0\}$ , we have $\gamma _{m,n}(a)\not =0$ for some large $n\ge m$ . Choose a function $g\in C([0, \lVert a\rVert )_+$ such that $0\le g(t)\le 1$ for all $t\in [0, \infty )$ , $g(t)=1$ if $t \in [\lVert a\rVert -\lVert a\rVert /2m, \lVert a\rVert )$ and $g(t)=0$ if $t\in [0,\lVert a\rVert -\lVert a\rVert /m]$ . We may assume that $\gamma _{m,n}(g(\lvert a\rvert ))\not =0$ for all $a\in {\cal F}_m\setminus \{0\}$ . It follows that $\left \lVert \gamma _{m,n}(\lvert a\rvert )\right \rVert \ge (1-1/m)\lVert a\rVert $ for all $a\in {\cal F}_m\setminus \{0\}$ . Thus we may assume that for any $a\in {\cal F}_m$ and all $n> m$ ,

(e6.5) $$ \begin{align} \left\lVert{\gamma}_{m,n}(a)\right\rVert\ge (1-1/m)\lVert a\rVert. \end{align} $$

In what follows, $A_{\mathfrak {d}}=\lim _{n\to \infty }\left (M_{d_n'}(B), \delta _n'\right )$ is the $C^*$ -algebra in Theorem 6.1 and $\delta _n'$ is as described in formula (e6.1) such that formula (e6.5) holds for $n\ge m+1$ .

We wish to construct a unital simple $C^*$ -algebra $A_z^C$ with a unique tracial state such that $K_0\left (A_z^C\right )=\mathbb {Z}$ and $K_1\left (A_z^C\right )=\{0\}$ .

The strategy is to have a Jiang–Su-style inductive limit of some $C^*$ -subalgebras of $C\!\left([0,1], M_p(B)\otimes M_q(B)\right )$ for some nonnuclear RFD algebra B, or perhaps some $C^*$ -subalgebra of $C\!\left([0,1], M_{pq}(B)\right )$ . However, there are several difficulties to be resolved. One should avoid using $M_p(B)\otimes M_q(B)$ as building blocks, since there are different $C^*$ -tensor products and potential difficulties in computing the K-theory. Other issues include the fact that each fibre $M_m(B)$ is not simple.

We begin with the following building blocks:

Definition 6.3. For a pair of integers $m,k\ge 1$ , define

$$ \begin{align*} E_{m,k}:= \{f\in C([0,1], M_{mk}(B)): f(0)\in M_{m}(B)\otimes 1_{k} \mathrm{\ and\ } f(1)\in 1_{m}\otimes M_{k}\}. \end{align*} $$

Note that here one views $M_m(B)\otimes 1_k, 1_m\otimes M_k \subset M_m(B)\otimes M_k = M_{mk}(B)$ as unital $C^*$ -subalgebras.

Fix integers $m,n\ge 1$ . Let $D(m,k)=M_m(B)\oplus M_k$ . Define $\varphi _0: D(m,k)\to M_m(B)\otimes 1_k$ by $\varphi _0((a,b)):= a\otimes 1_k$ for all $(a,b)\in D(m,k)$ and $\varphi _1: D(m,k)\to M_k$ by $\varphi _1((a,b))=1_m\otimes b$ .

Then

(e6.6) $$ \begin{align} E_{m,k}\cong \{(f,g)\in C([0,1], M_{mk}(B))\oplus D(m,k): f(0)=\varphi_0(g)\mathrm{\ and\ } f(1)=\varphi_1(g)\}. \end{align} $$

Denote by $\pi _e: E_{m,k} \to D(m,k)$ the quotient map which maps $(f,g)$ to g. Denote by $\pi _0: E_{m,k}\to M_m(B)\otimes 1_k$ the homomorphism defined by $\pi _0((f,g)):= \varphi _0(g)=f(0)$ and by $\pi _1: E_{m,k}\to 1_m\otimes M_k$ the homomorphism defined by $\pi _1((f,g)):= \varphi _1(g)=f(1)$ .

Lemma 6.4. If m and k are relatively prime, then $E_{m,k} $ has no proper projections.

Proof. Recall that $B=C_0 ((0,1], C)^{\sim }$ , the unitisation of $C_0((0,1], C)$ . Let $\tau _B$ be the tracial state on $M_m(B)$ induced by the quotient map $B\to B/C_0((0,1],C)\cong \mathbb {C}$ , and let $\mathrm {tr}_k$ be the tracial state of $M_k$ . Let $\tau =\tau _B\otimes \mathrm {tr}_k$ .

Let $e\in E_{m,k} $ be a nonzero projection. Note that $E_{m,k} \subset C([0,1], M_{mk}(B))$ . Note also that $K_0(B)=\mathbb {Z}$ and $1_B$ is the only nonzero projection of B. Then for each $x\in [0,1]$ , $e(x)$ is a nonzero projection in $M_{mk}(B)$ . One easily shows that $\tau (e(x))$ is a constant function on $[0,1]$ . Let $\tau (e(x))=r\in (0,1]$ . But $\tau (e(0))\in \{i/m: i=0,1,\dotsc ,m\}$ and $\tau (e(1))\in \{j/k,i=0,1,\dotsc ,k\}$ . Since m and k are relatively prime, $\tau (e(0))=\tau (e(1))=1$ . Hence $\tau (e(x))=1$ for all $x\in [0,1]$ . This is possible only when $e=1_m\otimes 1_k$ .

Lemma 6.5. Suppose that m and k are relatively prime. Then

$$ \begin{align*} \left(K_0\left(E_{m,k}\right), K_0\left(E_{m,k}\right)_+, \left[1_{E_{m,k}}\right]\right)=(\mathbb{Z}, \mathbb{N}\cup \{0\}, 1)\qquad \text{and} \qquad K_1\left(E_{m,k}\right)=\{0\}. \end{align*} $$

Proof. Let

$$ \begin{align*} I=\left\{f\in E_{m,k}: f(0)=f(1)=0\right\}. \end{align*} $$

Then $I\cong C_0((0,1))\otimes M_{mk}(B) =S(M_{mk}(B))$ . It follows that

(e6.7) $$ \begin{align} K_0(I)=K_1(M_{mk}(B))=\{0\}\qquad\text{and}\qquad K_1(I)=K_0(M_{mk}(B))=\mathbb{Z}. \end{align} $$

Consider the short exact sequence

(e6.8) $$ \begin{align} 0\to I\stackrel{\iota_I}{\longrightarrow} E_{m,k}\stackrel{\pi_e}{\longrightarrow} D(m,k)\to 0, \end{align} $$

where $\iota _I: I\to E_{m,k}$ is the embedding and $\pi _e: E_{m,k}\to D(m,k)$ is the quotient map. One obtains the following six-term exact sequence:

(e6.9) $$ \begin{align} \begin{array}{ccccc} K_0(I) & \stackrel{{\iota_I}_{*0}}\longrightarrow & K_0\left(E_{m,k}\right) & \stackrel{{\pi_e}_{*0}}{\longrightarrow} & K_0(D(m,k))\\ \uparrow_{\delta_1} & & & & \downarrow_{\delta_0}\\ K_1(D(m,k)) & \stackrel{{\pi_e}_{*1}}{\longleftarrow} & K_1\left(E_{m,k}\right) & \stackrel{{\iota_I}_{*1}}{\longleftarrow} & K_1(I), \end{array} \end{align} $$

which becomes

(e6.10) $$ \begin{align} \begin{array}{ccccc} 0 & \stackrel{{\iota_I}_{*0}}\longrightarrow & K_0\left(E_{m,k}\right) & \stackrel{{\pi_e}_{*0}}{\longrightarrow} & \mathbb{Z}\oplus \mathbb{Z}\\ \uparrow_{\delta_1} & & & & \downarrow_{\delta_0}\\ 0 & \stackrel{{\pi_e}_{*1}}{\longleftarrow} & K_1\left(E_{m,k}\right) & \stackrel{{\iota_I}_{*1}}{\longleftarrow} & \mathbb{Z}. \end{array} \end{align} $$

Note that

$$ \begin{align*} \mathrm{im}\left({\pi_e}_{*0}\right)=\left\{(x, y)\in K_0(D(m,k)): {\varphi_0}_{*0}(x)={\varphi_1}_{*0}(y)\right\}. \end{align*} $$

The lemma follows from a straightforward computation.

Set $\tau \in T(C([0,1], M_{mk}(B)))$ . By, for example, [26, Theorem 2.1] and the Choquet and Fubini theorems,

$$ \begin{align*} \tau(f)=\int_{\partial_e T\left(C\left(\left[0,1\right]\right)\right)\times \partial_eT\left(M_{mk}(B)\right)} f d(\mu\times\mu_B) \end{align*} $$

for all $f\in \operatorname {Aff}(T(C([0,1], M_{mk}(B))))_{\mathrm {sa}}$ , where $\mu $ is a probability Borel measure on $[0,1]$ and $\mu _B$ is a probability Borel measure on $\partial _eT(M_{mk}(B))$ . By the Fubini theorem again, we may write $\tau (f)=\int _{\left [0,1\right ]} \sigma _t(f(t))d\mu $ , where $\sigma _t$ is a tracial state of $M_{mk}(B)$ . Let I be the ideal in the proof of Lemma 6.5. Then $I\cong C_0((0,1))\otimes M_{mk}(B)$ . Now set $\tau \in T\left (E_{m,k}\right )$ such that $\lVert \tau \rvert _I\rVert \not =0$ . Since $(1/\lVert \tau _I\rVert )\tau \rvert _I$ can be extended to a tracial state of $C([0,1], M_{mk}(B))$ , we may write $\tau \rvert _{I}(f)=\int _{\left (0,1\right )} \sigma _t(f(t))d\mu $ for all $f\in C_0((0,1))\otimes M_{mk}(B)$ , where $\sigma _t$ is a tracial state of $M_{mk}(B)$ and $\mu $ is a Borel measure on $(0,1)$ (with $\lVert \mu \rVert =\lVert \tau \rvert _I\rVert \le 1$ ). Since $E_{m,k}/I=M_m(B)\oplus M_k$ , as in [31, 2.5], one may write

(e6.11) $$ \begin{align} \tau(f)=\int_0^1 \sigma_t(f(t))d\nu\mathrm{\ for\ all\ } f\in E_{m,k}, \end{align} $$

where $\sigma _0$ is a tracial state on $M_m(B)$ , $\sigma _1$ is a tracial state on $M_k$ , $\nu $ is a probability Borel measure on $[0,1]$ , $\nu \rvert _{\left (0,1\right )}=\mu \rvert _{\left (0,1\right )}$ , and if $\lVert \tau \rvert _I\rVert =0$ , then $\nu \rvert _{\left (0,1\right )}=0$ .

Notation 6.6. Let $\gamma : B\to M_r$ be a finite-dimensional representation with rank r – that is, $\gamma $ is a finite direct sum of irreducible representations $\gamma _j: j=1,2,\dotsc ,l$ , each of which has rank $r_j$ ( $1\le j\le l$ ), such that $r=\sum _{j=1}^l r_j$ . We will also use $\gamma $ for $\gamma \otimes \operatorname {id}_m:M_m(B)\to M_{rm}$ for all integers $m\ge 1$ . In what follows we may also write $M_L$ for $M_L(\mathbb {C}\cdot 1_B)$ for all integers $L\ge 1$ . In this way, $\gamma $ (or $\gamma \otimes \operatorname {id}_m$ ) is a homomorphism from $M_m(B)$ into $M_{rm}\subset M_{rm}(B)$ .

Let $\xi _0, \xi _1,\xi _2,\dotsc ,\xi _{{k-1}}: [0,1]\to [0,1]$ be continuous paths. Define a homomorphism

$$ \begin{align*} H: C([0,1], M_{mn}(B))\to C\left([0,1], M_{((k-1)r+1)mn}(B)\right) \end{align*} $$

by

$$ \begin{align*} H(f)(t):= \begin{pmatrix} f\circ \xi_0(t) & 0 &\dotsb & 0\\ 0 & \gamma(f\circ \xi_1 (t) ) &\dotsb &0\\ \vdots & \vdots & & \vdots\\ 0 & 0 & \dotsb & \gamma (f\circ \xi_{k-1}(t) )\end{pmatrix} \mathrm{\ for\ all\ } f\in C([0,1], M_{mn}(B)) \end{align*} $$

and $t\in [0,1]$ . Note that H can be also defined on $E_{m,n}\subset C([0,1], M_{mn}(B))$ . But in general, H does not map $E_{m,n}$ into $E_{m,n}$ . However, with some restrictions on the boundary (restriction on $\xi _i$ s), it is possible that H maps $E_{m,n}$ into $E_{m,n}$ .

For the convenience of the construction, let us add some notation and terminology.

Set $f, g\in M_n(B)$ . We write $f=^s g$ if there is a scalar unitary $w\in M_n$ such that $w^*fw=g$ . Also, if $f, g\in C([0,1], M_n(B))$ , we write $f=^s g$ if there is a unitary $w\in C([0,1], M_n)$ such that $w^*fw=g$ .

6.7. We will construct $A=\lim _{n\to \infty }(A_n, \varphi _m)$ . The construction will be by induction. Fix B as in Notation 6.6. Set $A_1=E_{3, 5}$ .

Denote by ${\bar {\mathfrak {3}}}$ the supernatural number $3^\infty $ . Write $A_{{\bar {\mathfrak {3}}}}=\lim _{n\to \infty } \left (M_{d_n'}(B), \delta _n'\right )$ (see Theorem 6.1), where

(e6.12) $$ \begin{align} \delta_n'(f)=\begin{pmatrix} f & 0\\ 0 & \gamma_n(f)\end{pmatrix} \mathrm{\ for\ all\ } f\in M_{d_n'}(B), \end{align} $$

as in formula (e6.1), which also has the properties in equations (e6.2) and (e6.3) (with $d_n=3^l$ for some integer $l\ge 1$ ). Hence, without loss of generality, by passing to a subsequence we may assume, for all n,

(e6.13) $$ \begin{align} \frac{1}{d_n-1} <1/3^n. \end{align} $$

Recall that $B=C_0((0,1], C)^{\sim }$ . For each $t\in [0,1]$ , denote by $\theta _t: B\to B$ the homomorphism defined, for all $f\in B$ , by

(e6.14) $$ \begin{align} \theta_t(f)(x) := f((1-t)x)\mathrm{\ for\ all\ } x\in (0,1]. \end{align} $$

Note also that for any integer $l\ge 1$ , we will use $\theta _t$ for $\theta _t\otimes \operatorname {id}_{l}: M_{l}(B)\to M_{l}(B)$ . Thus, if $f\in M_l(B)$ ,

(e6.15) $$ \begin{align} \theta_1(f)=f(0)\in M_l. \end{align} $$

It should be noted that $\theta _0=\operatorname {id}_{M_l(B)}$ .

We state the inductive step as the following lemma:

Lemma 6.8. For $A_m=E_{p_m, q_m}$ with $(p_m, q_m)=1$ , we have $(5, p_m)=1$ and $(3, q_m)=1$ . There exist $A_{m+1}=E_{p_{m+1}, q_{m+1}}$ , where $(p_{m+1}, q_{m+1})=1$ , $(5, p_{m+1})=1$ and $(3, q_{m+1})=1$ , and a unital injective homomorphism $\varphi _m: A_m\to A_{m+1}$ of the form

(e6.16) $$ \begin{align} \varphi_{m}(f) (t) =u^*\begin{pmatrix} \Theta_{m}(f)(t) & 0 &\dotsb & 0\\ 0 & \gamma_{m}(f\circ \xi_1(t))\otimes 1_5 &\dotsb &0\\ \vdots & \vdots & & \vdots\\ 0 & 0 & \dotsb & \gamma_m(f\circ \xi_k(t))\otimes 1_5 \end{pmatrix} u \end{align} $$

for all $f\in A_m$ , where $u\in U\left (C\!\left([0,1], M_{p_{m+1}q_{m+1}}\right )\right )$ , $\Theta _{m}: A_m\to C\!\left([0,1], M_{5kp_mq_m}(B)\right )$ is a homomorphism, $k\ge 1$ is an integer, $t\in [0,1]$ and $\gamma _m: M_{p_mq_m}(B)\to M_{R(m)p_mq_m}$ is a finite-dimensional representation, where $R(m)\ge 1$ is an integer. Moreover, the following are true:

1. (1) Each $\xi _i: [0,1]\to [0,1]$ is a continuous map which has one of the following three forms:

   (e6.17) $$ \begin{align} \xi_i(t)&=\begin{cases} (2/3)t & \text{if } t\in [0,3/4],\\ 1/2 & \text{if } t\in (3/4,1],\end{cases} \hspace{2pc} \end{align} $$
   (e6.18) $$ \begin{align} \xi_i(t)&=1/2\mathrm{\ for\ all\ } t\in [0,1], \hspace{3.5pc} \end{align} $$
   (e6.19) $$ \begin{align} \xi_i(t)&=\begin{cases} 1/2+(2/3)t & \text{if } t\in [0,3/4],\\ 1 & \text{if } t\in (3/4,1], \end{cases} \end{align} $$
   and each type of $\xi _i$ appears in equation (e6.16) at least once.

2. (2) $5k/5kR(m)= 1/R(m) <1/3^{m}$ .

3. (3) For a fixed finite subset ${\cal F}_m\subset A_m\setminus \{0\}\subset C\!\left([0,1], M_{p_mq_m}(B)\right )$ ,

   $$ \begin{align*} \lVert\gamma_m(f(t))\rVert> (1-1/2m)\lVert f\rVert \not=0\mathrm{\ for\ some\ }t \in [0,1]. \end{align*} $$

4. (4) We have

   (e6.20) $$ \begin{align} \Theta_m(f)=\mathrm{diag}\left(\theta^{(1)}(f) ,\dotsc, \theta^{(k)}(f)\right), \end{align} $$
   where $\theta ^{(i)}: E_{p_m, q_m}\to C\!\left([0,1], M_{5p_mq_m}(B)\right )$ is defined – if $\xi _i(3/4)=1/2$ – for each $f\in E_{p_m, q_m}$ by
   $$ \begin{align*} \theta^{(i)}(f)(t):= \begin{cases} f(\xi_i(t)) & \text{if }t\in [0,3/4],\\ \theta_{4\left(t-3/4\right)}(f(\xi_i(t))) & \text{if }t\in (3/4,1], \end{cases} \end{align*} $$
   and where $\theta _t: M_{p_mq_m}(B)\to M_{p_mq_m}(B)$ (recall that $B= C_0((0,1], C)^{\sim } $ ) is a unital homomorphism defined by
   (e6.21) $$ \begin{align} \theta_t(f)(x):= f((1-t)x)\mathrm{\ for\ all\ } x\in (0,1] {\mathrm{\ and\ all\ }} t\in [0,1] \end{align} $$
   and, if $\xi _i(3/4)=1$ , for each $f\in E_{p_m,q_m}$ ,
   $$ \begin{align*} \theta^{(i)}(f)(t)=f(\xi_i(t)),\quad{{t\in [0,1].}} \end{align*} $$

Proof. To avoid the potential complication of computing relative primality of integers, we will have a three-stage construction.

Stage 1: Write $A_m=E_{p_m, q_m}$ , where $(p_m, q_m)=1$ . Also $(5, p_m)=1$ and $(3, q_m)=1$ .

Fix any finite subset ${\cal F}_m\subset E_{p_m,q_m}\setminus \{0\}$ . One can choose a finite subset $S\subset [0,1]$ such that, for any $f\in {\cal F}_m$ , there is $s\in S$ , $ \lVert f(s)\rVert> (1-1/2m)\lVert f\rVert \not =0. $ Note that ${\cal F}'=\{f(s): f\in {\cal F}_m\text { and } s\in S\}\setminus \{0\}$ is a finite subset of $M_{p_mq_m}(B)$ . By passing to a subsequence, we may assume (replacing $\gamma _m$ by ${\gamma }_{m,n}$ as mentioned in formula (e6.5)) that

(e6.22) $$ \begin{align} \lVert\gamma_m(g)\rVert>(1-1/2m)\lVert g\rVert\not=0 \mathrm{\ for\ all\ } g\in {\cal F}'. \end{align} $$

It follows that for any $f\in {\cal F}_m$ ,

(e6.23) $$ \begin{align} \lVert\gamma_m(f(s'))\rVert \ge (1-1/2m)\lVert f\rVert \not=0{\mathrm{\ for\ some\ }} s'\in S\subset [0,1]. \end{align} $$

Define $\psi _m': M_{p_m}(B)\otimes M_{q_m}\to M_{d_mp_m}(B)\otimes M_{5q_m}$ by $ \psi _m':= \delta _m\otimes s$ , where

(e6.24) $$ \begin{align} \delta_{m}(a)=\begin{pmatrix} a & 0\\ 0 & \gamma_m(a)\end{pmatrix}\mathrm{\ for\ all\ } a\in M_{p_m}(B),\mathrm{\ and\ } s(c)=c\otimes 1_5\mathrm{\ for\ all\ } c\in M_{q_m}. \end{align} $$

Define $\psi _m: E_{p_m, q_m}\to E_{d_{m} p_m, 5q_m}$ by

(e6.25) $$ \begin{align} \psi_m(f)(t):= \psi_m'(f(t))\mathrm{\ for\ all\ } f\in E_{p_m,q_m}\mathrm{\ and\ } t\in [0,1]. \end{align} $$

Set $f\in E_{p_m, q_m}$ . Then $f(0)=b\otimes 1_{q_m}$ , where $b\in M_{p_m}(B)$ . Thus,

(e6.26) $$ \begin{align} \psi_m(f)(0)=\psi_m'(f(0))=\delta_m (b)\otimes \left(1_{q_m}\otimes 1_5\right)\in M_{d_mp_m}(B)\otimes 1_{5q_m}. \end{align} $$

On the other hand, $f(1)=1_{p_m}\otimes c$ , where $c\in M_{q_m}$ . Thus

(e6.27) $$ \begin{align} \psi_m(f)(1)=\psi_m'(f(1))=1_{d_mp_m}\otimes (c\otimes 1_5)\in 1_{d_mp_m}\otimes M_{5q_m}. \end{align} $$

So indeed, $\psi _m$ maps $E_{p_m, q_m}$ into $E_{d_mp_m, 5q_m}$ .

Note that for $t\in [0,1]$ , we have for all $f\in E_{p_m,q_m}$ (writing $\gamma _m$ for $\gamma _m\otimes \operatorname {id}_{M_{q_{_m}}}$ )

(e6.28) $$ \begin{align} \psi_m(f)(t)=\psi_m'(f(t))=\begin{pmatrix} f(t) & 0\\ 0 & \gamma_m(f(t))\end{pmatrix} \otimes 1_5. \end{align} $$

Recall that $\gamma _m: M_{p_mq_m}(B)\to M_{R(m)p_mq_m}$ is a unital homomorphism with $R(m)=d_m-1$ . Note that by formula (e6.13), we may assume that $R(m)>3^m$ .

Stage 2: We will use a modified construction of Jiang and Su and define $\varphi _m$ on $[0,3/4]$ .

Choose a (first) pair of different prime numbers $k_0$ and $k_1$ such that

(e6.29) $$ \begin{align} k_0> 15 q_m\qquad\text{and}\qquad k_1>15 k_0d_mp_m. \end{align} $$

In particular, $k_0, k_1\not =3, 5$ .

Recall that $(3,q_m)=1$ , $(5,p_m)=1$ and $d_m=3^{l_m}$ for some $l_m\ge 1$ . Therefore, $(k_0d_mp_m, k_15q_m)=1$ . Let $p_{m+1}=k_0d_mp_m$ , $q_{m+1}=k_15q_m$ and $k=k_0k_1$ . Then $(p_{m+1}, q_{m+1})=1$ , $(5, p_{m+1})=1$ and $(3, q_{m+1})=1$ . Write

(e6.30) $$ \begin{align} k=r_0+m(0)q_{m+1}\qquad\text{and}\qquad k=r_1+m(1)p_{m+1}, \end{align} $$

where $m(0), r_0, m(1), r_1\ge 1$ are integers and

(e6.31) $$ \begin{align} 0<r_0<q_{m+1}, \quad r_0\equiv k(\bmod q_{m+1}), \end{align} $$

(e6.32) $$ \begin{align} 0<r_1< p_{m+1}, \quad r_1\equiv k (\bmod p_{m+1}). \end{align} $$

Moreover, by formula (e6.29),

(e6.33) $$ \begin{align}\nonumber k-r_1-r_0&>k-q_{m+1}-p_{m+1}=k-k_15q_m-k_0d_mp_m\\\nonumber &= k_1\left(k_0-5q_m\right)-k_0d_mp_m \\ &>k_1\left(10q_m\right)-k_0d_mp_m> 0. \end{align} $$

We will construct paths $\xi _i$ . At $t=0$ , define

(e6.34) $$ \begin{align} \xi_i(0):= \begin{cases} 0 & {\text{if }}1\le i\le r_0,\\ 1/2 & \text{if } r_0<i\le k. \end{cases} \end{align} $$

Note that since

(e6.35) $$ \begin{align} r_05q_m\equiv k5q_m\equiv k_0k_15q_m\equiv 0 (\bmod q_{m+1}), \end{align} $$

$r_05q_m=t_0q_{m+1}$ for some integer $t_0\ge 1$ . Note also that if $f\in E_{d_mp_m, 5q_m}$ , then $f(0)=b\otimes 1_{5q_m}$ for some $b\in M_{d_mp_m}(B)$ . Hence $f(0)\otimes 1_{r_0}=b\otimes 1_{r_05q_m}=\left (b\otimes 1_{t_0}\right )\otimes 1_{q_{m+1}}$ for any $f\in E_{d_mp_m, 5q_m}$ . On the other hand, for any $f\in E_{d_mp_m,5q_m} $ ,

(e6.36) $$ \begin{align} \mathrm{diag}\left(f\left(\xi_{r_0+1}(0)\right), \dotsc, f(\xi_k(0))\right)=^s(f(1/2))\otimes 1_{m(0)q_{m+1}}. \end{align} $$

In fact, there is a scalar unitary $s_0\in M_{m(0)q_{m+1}d_mp_m5q_m}$ such that

$$ \begin{align*} s_0^*\mathrm{diag}(\overbrace{b,b,\dotsc,b}^{k-r_0})s_0=b\otimes 1_{m(0)q_{m+1}}\mathrm{\ for\ all\ } b\in M_{d_mp_m5q_{m}}(B) \end{align*} $$

(recall that $f(1/2)\in M_{d_mp_m5q_{m}}(B)$ ). Therefore, there exists a unitary $v_0\in U\left (M_{p_{m+1}q_{m+1}}\right )$ such that for all $f\in E_{d_mp_m,5q_m}$ ,

(e6.37) $$ \begin{align} \rho_0(f):=v_0^*\begin{pmatrix} f(\xi_1(0)) & 0 &\dotsb & 0\\ 0 & f(\xi_2(0)) &\dotsb &0\\ \vdots & \vdots & & \vdots\\ 0 & 0 & \dotsb & f( \xi_{k}(0))\end{pmatrix}v_0 \end{align} $$

is in $M_{p_{m+1}}(B)\otimes 1_{q_{m+1}}$ . So $\rho _0$ defines a homomorphism from $E_{d_mp_m, 5q_m}$ into $M_{p_{m+1}}(B)\otimes 1_{q_{m+1}}$ .

At $t=3/4$ , define

(e6.38) $$ \begin{align} \xi_i(3/4):= \begin{cases} 1/2 & \text{if } 1\le i\le k-r_1,\\ 1 & \text{if }k-r_1<i\le k. \end{cases} \end{align} $$

As in the case at $0$ , by formula (e6.32),

$$ \begin{align*} r_1d_mp_m\equiv kd_mp_m\equiv k_0k_1d_mp_m\equiv 0 (\bmod p_{m+1}). \end{align*} $$

So one may write $r_1d_mp_m=t_1p_{m+1}$ for some integer $t_1\ge 1$ . Set $f\in E_{d_mp_m, 5q_m}$ . Then $f(1)=1_{d_mp_m}\otimes c$ for some $c\in M_{5q_m}$ . It follows that $1_{r_1}\otimes f(1)=1_{p_{m+1}}\otimes \left (1_{t_1}\otimes c\right )$ . Also,

$$ \begin{align*} \mathrm{diag}\left(f(\xi_1(3/4)), \dotsc, f\left(\xi_{k-r_1}(3/4)\right)\right)=^s1_{m(1)p_{m+1}}\otimes f(1/2). \end{align*} $$

In fact, there is a scalar unitary $s_1\in M_{m(1)p_{m+1}d_mp_m5q_m}$ such that

$$ \begin{align*} s_1^*\mathrm{diag}(\overbrace{b, b,\dotsc,b}^{k-r_1})s_1=1_{m(1)p_{m+1}}\otimes b\mathrm{\ for\ all\ } b\in M_{d_mp_m5q_m}(B). \end{align*} $$

Thus there is a unitary $v_{3/4}\in U\left (M_{p_{m+1}q_{m+1}}\right )$ such that for $f\in E_{d_mp_m, 5q_m}$ ,

(e6.39) $$ \begin{align} \rho_{3/4}(f) :=v_{3/4}^*\begin{pmatrix} f(\xi_1(3/4)) & 0 &\dotsb & 0\\ 0 & f(\xi_2(3/4)) &\dotsb &0\\ \vdots & \vdots & & \vdots\\ 0 & 0 & \dotsb & f(\xi_{k}(3/4))\end{pmatrix}v_{3/4} \end{align} $$

defines a homomorphism from $E_{d_mp_m, 5q_m}$ to $1_{p_{m+1}}\otimes M_{q_{m+1}}(B)$ .

To connect $\xi _i(0)$ and $\xi _i(3/4)$ continuously, on $[0,3/4]$ , let us define (see formula (e6.33))

(e6.40) $$ \begin{align} \xi_i(t):= \begin{cases} 2t/3 & \text{if } 1\le i\le r_0,\\ 1/2 & \text{if } r_0<i\le k-r_1,\\ 1/2+2t/3 &\text{if } k-r_1<i\le k.\end{cases} \end{align} $$

Let $v\in C\!\left([0, 3/4], M_{p_{m+1}q_{m+1}}\right )$ be a unitary such that $v(0)=v_0$ and $v(3/4)=v_{3/4}$ . Now, on $[0,3/4]$ , define, for all $f\in E_{p_m, q_m}$ ,

(e6.41) $$ \begin{align} \varphi_m(f)(t):= v(t)^*\begin{pmatrix} \psi_m(f)\circ\xi_1(t) & 0 &\dotsb & 0\\ 0 & \psi_m(f)\circ \xi_2(t) &\dotsb &0\\ \vdots & \vdots & & \vdots\\ 0 & 0 & \dotsb & \psi_m(f)\circ \xi_{k}(t)\end{pmatrix}v(t). \end{align} $$

Stage 3: We connect $3/4$ to $1$ , recalling that $\pi _1\left (E_{p_{m+1}, q_{m+1}}\right )=1_{p_{m+1}}\otimes M_{q_{m+1}}$ .

We first extend $\xi _i$ by defining

(e6.42) $$ \begin{align} \xi_i(t)=\xi_i(3/4) \mathrm{\ for\ all\ } t\in (3/4,1],\quad i=1,2,\dotsc,k. \end{align} $$

Recall equation (e6.28); at $t=3/4$ , for each i and for $f\in E_{p_m,q_m}$ ,

(e6.43) $$ \begin{align} \psi_m(f)(\xi_i(3/4))=\begin{pmatrix} f(\xi_i(3/4)) & 0\\ 0 & \gamma_m(f(\xi_i(3/4)))\end{pmatrix} \otimes 1_5. \end{align} $$

For $k-r_1< i\le k$ , define, for $t\in ( 3/4, 1]$ and $f\in E_{p_m,q_m}$ ,

(e6.44) $$ \begin{align} \hspace{-0.2in} {\tilde \psi}_{m,i}(f)(t) := \begin{pmatrix} f(\xi_i(3/4)) & 0\\ 0& \gamma_m(f(\xi_i(3/4))) \end{pmatrix}\otimes 1_5 = \begin{pmatrix} f(1) & 0\\ 0& \gamma_m(f(1))\end{pmatrix}\otimes 1_5. \end{align} $$

Recall that $\gamma _m(f(1))=f(1)\otimes 1_{d_m-1}$ . Therefore there exists a scalar unitary $s_3\in M_{p_mq_m5d_m}$ such that

$$ \begin{align*} s_3^*\left({\tilde \psi}_{m,i}(f)(t)\right)s_3=1_{d_m}\otimes f(1)\otimes 1_5 \mathrm{\ for\ all\ } f\in E_{p_m,q_m},\quad t\in [3/4,1]. \end{align*} $$

Note that $f(1)$ has the form $1_{p_m}\otimes c$ for some $c\in M_{q_m}$ . So there is a scalar unitary $s_4\in M_{t_1p_{m+1}q_m 5}$ such that

(e6.45) $$ \begin{align} s_4^*\mathrm{diag}\left({\tilde \psi}_{m,k-r_1+1}(f)(t), \dotsc, {\tilde \psi}_{m,k}(f)(t)\right) s_4= 1_{r_1d_mp_m} \otimes c \otimes 1_5 = 1_{t_1p_{m+1}} \otimes c \otimes 1_5. \end{align} $$

Now recall formula (e6.14) for the definition of $\theta _t$ . For $1\le i\le k-r_1$ , define, for $t\in (3/4, 1]$ ,

(e6.46) $$ \begin{align} {\tilde \psi}_{m,i}(f)(t):= \begin{pmatrix}\theta_{4\left(t-3/4\right)}(f(\xi_i(t))) & 0\\ 0& \gamma_m(f(\xi_i(t)))\end{pmatrix}\otimes 1_5\mathrm{\ for\ all\ } f\in E_{p_m, q_m}. \end{align} $$

Note that $\theta _1(f(\xi _i(3/4)))=\theta _1(f(1/2))\in M_{p_mq_m}$ (see formulas (e6.38) and (e6.15)). Recall that $\varphi _m(f)(3/4)\in 1_{p_{m+1}}\otimes M_{q_{m+1}}(B)$ . Note that for $1\le i\le k-r_1$ ,

$$ \begin{align*} {\tilde \psi}_{m,i}(f)(1)=\begin{pmatrix}\theta_1(f(1/2)) & 0\\ 0& \gamma_m(f(1/2))\end{pmatrix}\otimes 1_5\mathrm{\ for\ all\ } f\in E_{p_m, q_m}. \end{align*} $$

Moreover (see also equation (e6.30)), there is a scalar unitary $s_5\in M_{m(1)p_{m+1}d_mp_mq_m 5}$ such that

$$ \begin{align*} s_5^*\mathrm{diag}\left({\tilde \psi}_{m,1} (f)(1), \dotsc, {\tilde \psi}_{m, k-r_1} (f)(1)\right)s_5=1_{m(1)p_{m+1}}\otimes \begin{pmatrix} \theta_1(f(1/2)) & 0\\ 0& \gamma_m(f(1/2))\end{pmatrix}\otimes 1_5. \end{align*} $$

Thus, for $t=1$ , there is a unitary $v_1\in 1_{p_{m+1}}\otimes M_{q_{m+1}}$ such that

(e6.47) $$ \begin{align} \rho_1(f):= v_1^*\begin{pmatrix} {\tilde \psi}_{m,1}(f)(1) & 0 &\dotsb & 0\\ 0 & {\tilde \psi}_{m,2}(f)(1) &\dotsb &0\\ \vdots & \vdots & & \vdots\\ 0 & 0 & \dotsb & {\tilde \psi}_{m,k}(f)(1)\end{pmatrix}v_1 \end{align} $$

defines a homomorphism from $E_{p_m, q_m}$ to $1_{p_{m+1}}\otimes M_{q_{m+1}}$ . There is a continuous path of unitaries $\{v(t): t\in [3/4, 1]\}\subset M_{p_{m+1}q_{m+1}}$ such that $v(3/4)$ is as defined and $v(1)=v_1$ – so now $v\in C\!\left([0,1], M_{p_{m+1}q_{m+1}}\right )$ with $v(0)=v_0$ and $v(1)=v_1$ , and $v(3/4)$ is consistent with the previous definition. Now define, for $t\in (3/4, 1]$ ,

(e6.48) $$ \begin{align} \varphi_m(f)(t):= v(t)^*\begin{pmatrix} {\tilde \psi}_{m,1}(f)(t) & 0 &\dotsb & 0\\ 0 & {\tilde \psi}_{m,2}(f)(t) &\dotsb &0\\ \vdots & \vdots & & \vdots\\ 0 & 0 & \dotsb & {\tilde \psi}_{m,k}(f)(t)\end{pmatrix}v(t) \end{align} $$

for all $f\in E_{p_m, q_m}$ . Note that by formulas (e6.43), (e6.46), (e6.47) and (e6.39),

(e6.49) $$ \begin{align} \varphi_m(f)(1)=\rho_1(f)\mathrm{\ for\ all\ } f\in E_{p_m,q_m}. \end{align} $$

Hence $\varphi _m$ is a unital injective homomorphism from $E_{p_m, q_m}$ to $E_{p_{m+1}, q_{m+1}}$ . (Note that injectivity follows from the fact that $\cup _{i=1}^k \xi _i([0,1])=[0,1]$ , as $r_0\ge 1$ and $k-r_1>0$ .)

For convenience of notation and for later use, let us define $\tilde \psi _{m,i}: E_{p_m, q_m} \to C\!\left([0,1], M_{d_mp_m5q_m}(B)\right )$ by

(e6.50) $$ \begin{align} \tilde \psi_{m,i}(f)(t):= \begin{cases} \psi_m(f\circ \xi_i(t)) &\text{if } t\in [0,3/4],\\ \tilde \psi_{m,i}(f)(t) &\text{if } t\in (3/4,1], \end{cases} \end{align} $$

for all $f\in E_{p_m,q_m}$ . Then we may write, for all $t\in [0,1]$ and all $f\in E_{p_m, q_m}$ ,

(e6.51) $$ \begin{align} \varphi_m(f)(t)=v(t)^*\begin{pmatrix} {\tilde \psi}_{m,1}(f)(t) & 0 &\dotsb & 0\\ 0 & {\tilde \psi}_{m,2}(f)(t) &\dotsb &0\\ \vdots & \vdots & & \vdots\\ 0 & 0 & \dotsb & {\tilde \psi}_{m,k}(f)(t)\end{pmatrix}v(t). \end{align} $$

Define $\theta ^{(i)'}: E_{p_m, q_m}\to C\!\left([0,1], M_{p_mq_m}(B)\right )$ , for each $f\in E_{p_m, q_m}$ , by

(e6.52) $$ \begin{align} \theta^{(i)'}(f)(t):= \begin{cases} f(\xi_i(t)) & \text{if }t\in [0,3/4],\\ \theta_{4\left(t-3/4\right)}(f(\xi_i(t))) & \text{if }t\in (3/4,1],\end{cases} \end{align} $$

if $\xi _i(3/4)=1/2;$ and if $\xi _i(3/4)=1$ , define

(e6.53) $$ \begin{align} \theta^{(i)'}(f)(t):= f(\xi_i(t))\mathrm{\ for\ all\ } t\in [0,1]. \end{align} $$

Define $\theta ^{(i)}(f):=\theta ^{(i)'}(f)\otimes 1_5$ for $f\in E_{p_m,q_m}$ and $\Theta _m: E_{p_m, q_m}\to C\!\left([0,1], M_{5kp_{m}q_{m}}(B)\right )$ , for each $f\in E_{p_m, q_m}$ , by

(e6.54) $$ \begin{align} \Theta_m(f):= \mathrm{diag}\left(\theta^{(1)}(f) ,\dotsc, \theta^{(k)}(f)\right)\mathrm{\ for\ all\ } t\in [0,1]. \end{align} $$

By formulas (e6.48), (e6.28) and (e6.46), as well as the definition of $\Theta _m$ , and by conjugating another unitary in $C\!\left([0,1], M_{p_{m+1}q_{m+1}}\right )$ , we may write

(e6.55) $$ \begin{align} \varphi_{m}(f)=u^*\begin{pmatrix} \Theta_{m}(f) & 0 &\dotsb & 0\\ 0 & \gamma_{m}(f\circ \xi_1) \otimes 1_5 &\dotsb &0\\ \vdots & \vdots & & \vdots\\ 0 & 0 & \dotsb & \gamma_m(f\circ \xi_k)\otimes 1_5 \end{pmatrix} u \mathrm{\ for\ all\ } f\in A_m. \end{align} $$

So $\varphi _m$ does have the form of formula (e6.16). Condition (1) of the lemma follows from the definition of $\xi _i$ and formulas (e6.40), (e6.42) and (e6.33). Condition (2) follows from formulas (e6.13) and (e6.28) (and two lines after it). Condition (3) follows from formula (e6.23). Finally, condition (4) follows from the definition of $\Theta _m$ .

Definition 6.9. From 6.7 and Lemma 6.8, inductively, we define $A_1=E_{3,5}$ , $A_m=E_{p_m, q_m}$ and homomorphism s $\varphi _m: A_m\to A_{m+1}=A_{p_{m+1}, q_{m+1}}$ as described in Lemma 6.8. Then we define $A=\lim _{n\to \infty }(A_m, \varphi _m)$ .

Remark 6.10. It should be noted that if $f(0), f(1)\in M_{p_mq_m}$ , then $\Theta _m(f(0))$ and $\Theta _m(f(1))$ are also scalar matrices.

7. Conclusion of the construction

Definition 7.1. Let $\{\xi _i: 1\le i\le m\}$ be a collection of maps described in Lemma 6.8(1). Note that each of the three types occurs at least once. Such a collection is said to be full. Let $C_2:=\left \{\xi _i^{(1)}\circ \xi _j^{(2)}: 1\le i\le m_1, 1\le j\le m_2\right \}$ be a collection of compositions of two maps in Lemma 6.8(1). This collection is called full if $\left \{\xi _{j}^{(2)}: 1\le j\le m_2\right \}$ is a full collection, and for each fixed $\xi _j^{(2)}$ , $\left \{\xi _i^{(1)}: \xi _i^{(1)}\circ \xi _j^{(2)}\in C_2\right \}$ is also a full collection. Inductively, a collection of n compositions of maps in Lemma 6.8(1),

$$ \begin{align*} C_n=\left\{\xi_{j(1)}^{(1)}\circ \xi_{j(2)}^{(2)}\circ \dotsb \circ \xi_{j(n)}^{(n)}: 1\le j(i)\le m_i: i=1,2,\dotsc,n\right\} \end{align*} $$

is called full if $\left \{\xi _{j(n)}^{(n)}: 1\le j(n)\le m_n\right \}$ is a full collection, and for each fixed $\xi _{j(n)}^{(n)}$ , the collection

$$ \begin{align*} \left\{\xi_{j(1)}^{(1)}\circ \xi_{j(2)}^{(2)}\circ\dotsb \circ \xi_{j(n-1)}^{(n-1)}: \xi_{j(1)}\circ \xi_{j(2)}\circ\dotsb \circ \xi_{j(n-1)}\circ \xi_{j(n)}\in C_n\right\} \end{align*} $$

is full.

Lemma 7.2. Let $\Xi =\xi _{j(1)}\circ \xi _{j(2)}\circ \dotsb \circ \xi _{j(n)}: [0,1]\to [0,1]$ be a composition of n maps, where each $\xi _{j(k)}: [0,1]\to [0,1]$ ( $1\le k\le n$ ) is one of the three types of continuous maps given in Lemma 6.8(1). Then for any $x, y\in [0,1]$ ,

(e7.1) $$ \begin{align} \lvert\Xi(x)-\Xi(y)\rvert \le (2/3)^n. \end{align} $$

Moreover, if $\left \{\Xi _j: 1\le j\le l\right \}$ is a full collection of compositions of n maps as before, then

(e7.2) $$ \begin{align} \cup^l_{j=1} \Xi_j([0,1])=[0,1], \end{align} $$

and for each $t\in [0,1]$ , $ \left \{\Xi _j(t): 1\le j\le l\right \} $ is $(2/3)^n$ -dense in $[0,1]$ .

Proof. Note that for each i and for any $x,y\in [0,1]$ , we have $\lvert \xi _i(x)-\xi _i(y)\rvert \leq (2/3)\lvert x-y\rvert $ . Then, by induction, for all $x, y\in [0,1]$ ,

(e7.3) $$ \begin{align} \lvert\Xi(x)-\Xi(y)\rvert &= \left\lvert\xi_{j(1)}\circ \xi_{j(2)}\circ \dotsb \circ \xi_{j(n)}(x)- \xi_{j(1)}\circ \xi_{j(2)}\circ \dotsb \circ \xi_{j(n)}(y)\right\rvert \end{align} $$

(e7.4) $$ \begin{align} &\le (2/3)\left\lvert\xi_{j(2)}\circ \dotsb \circ \xi_{j(n)}(x)- \xi_{j(2)}\circ \dotsb \circ \xi_{j(n)}(y)\right\rvert \hspace{3.5pc} \end{align} $$

(e7.5) $$ \begin{align} &\le\dotsb \le (2/3)^n\lvert x-y\rvert\le(2/3)^n. \hspace{10.5pc} \end{align} $$

One already observes that $\cup _{j\in S}\xi _j([0,1])=[0,1]$ if $\left \{\xi _j: j\in S\right \}$ is a full collection. An induction shows that if $\left \{\Xi _j: 1\le j\le l\right \}$ is a full collection, then

$$ \begin{align*} \cup_{j=1}^l \Xi_j([0,1])=[0,1]. \end{align*} $$

To show the last statement, fix $t\in [0,1]$ . Set $x\in [0,1]$ . Then for some $y\in [0,1]$ and $j\in \{1,2,\dotsc ,l\}$ ,

$$ \begin{align*} \Xi_j(y)=x. \end{align*} $$

Now, by the first part of the statement, for any $t\in [0,1]$ ,

$$ \begin{align*} \left\lvert\Xi_j(t)-\Xi_j(y)\right\rvert \le (2/3)^n. \end{align*} $$

It follows that

$$ \begin{align*} \left\lvert\Xi_j(t)-x\right\rvert=\left\lvert\Xi_j(t)-\Xi_j(y)\right\rvert \le (2/3)^n. \\[-38pt] \end{align*} $$

Theorem 7.3. The inductive limit A defined in Definition 6.9 can be made into a unital simple $C^*$ -algebra $A_z^C$ such that

(e7.6) $$ \begin{align} \left(K_0(A_z), K_0(A_z)_+, \left[1_{A_z}\right], K_1(A_z)\right)=(\mathbb{Z}, \mathbb{Z}_+, 1, \{0\}). \end{align} $$

If C is not exact, then $A_z^C$ is not exact.

Proof. For convenience, one makes an additional requirement in the construction. Let ${\cal F}_{m,1}\subset {\cal F}_{m,2},\dotsc ,{\cal F}_{m,n},\dotsc $ be an increasing sequence of finite subsets of $A_m$ such that $\cup _n {\cal F}_{m,n}$ is dense in $A_m$ .

One requires $\varphi _m\left ({\cal F}_{m,m+1} \right )\subset {\cal F}_{m+1,1}$ and $\varphi _m\left ({\cal F}_{m, m+n}\right )\subset {\cal F}_{m+1,n}$ , $m, n=1,2,\dotsc $ .

This is done inductively as follows: Choose any increasing sequence of finite subsets ${\cal F}_{1,1}\subset {\cal F}_{1,2},\dotsc , \subset A_1$ such that $\cup _n {\cal F}_{1,n}$ is dense in $A_1$ . Specify ${\cal F}_1={\cal F}_{1,1}\setminus \{0\}$ . Choose $A_2$ and define $\varphi _1: A_1\to A_2$ as in the construction of Lemma 6.8.

Choose an increasing sequence of finite subsets ${\cal F}_{2,1}, {\cal F}_{2,2},\dotsc $ of $A_2$ such that $\varphi _1\left ({\cal F}_{1,n}\right )\subset {\cal F}_{2,n}$ ( $n=1,2,\dotsc $ ) such that $\cup _n{\cal F}_{2,n}$ is dense in $A_2$ . Specify ${\cal F}_2={\cal F}_{2,1}\setminus \{0\}$ .

Once ${\cal F}_{m,1},{\cal F}_{m,2},\dotsc $ are determined, specify ${\cal F}_m={\cal F}_{m,1}\setminus \{0\}$ . Then construct $A_{m+1}$ and $\varphi _m: A_m\to A_{m+1}$ as in Lemma 6.8. Choose ${\cal F}_{m+1,1}, {\cal F}_{m+1,2},\dotsc $ so that $\varphi _m\left ({\cal F}_{m,m+1}\right )\subset {\cal F}_{m+1,1}$ and $\varphi _m\left ({\cal F}_{m,m+n}\right )\subset {\cal F}_{m+1,n}$ , as well as ${\cal F}_{m+1,n}\subset {\cal F}_{m+1,n+1}$ . Moreover, $\cup _n {\cal F}_{m+1,n}$ is dense in $A_{m+1}$ . Choose ${\cal F}_{m+1}={\cal F}_{m+1,1}\setminus \{0\}$ . Thus the requirement can be made.

Let us now prove that A is simple. For this, we will prove the following claim:

Claim: For any fixed i, and $g\in A_i\backslash \{0\}$ , there exists $n>i$ such that $\varphi _{i,n}(g)$ is full in $A_n$ . Without loss of generality, we may assume that $\lVert g\rVert =1$ . There are j and $f\in {\cal F}_{i,j}$ such that $\lVert f-g\rVert <1/64$ .

To simplify notation, without loss of generality we may write $i=1$ . Set $\varphi _{j,j'}=\varphi _{j'-1}\circ \dotsb \circ \varphi _j$ for $j'>j$ . Then $\varphi _{1, j'}(f) \in \varphi _{j'}\left ({\cal F}_{j'-1, j'}\right )\subset {\cal F}_{j', 1}$ . Recall also that each $\varphi _j$ is unital and injective. To further simplify the notation, without loss of generality we may write ${\cal F}_{i,j}\setminus \{0\}={\cal F}_m={\cal F}_{m,1}\setminus \{0\}$ . We assume that $m>128$ . By construction, for some $t\in (0,1)$ ,

(e7.7) $$ \begin{align} \lVert\gamma_m(f(t))\rVert> (1-1/2m)\lVert f\rVert\not=0. \end{align} $$

By continuity, there is $n(m)\ge 1$ such that for any $(2/3)^{n(m)-1}$ -dense set S of $[0,1]$ ,

(e7.8) $$ \begin{align} \lVert\gamma_m(f(s))\rVert \ge (1-1/2m)\lVert f\rVert\not=0\mathrm{\ for\ some\ }s \in S. \end{align} $$

For any $f\in C\!\left([0,1], M_{p_mq_m}(B)\right )$ and i, denote $h(t)=\gamma _m(f\circ \xi _i(t)){\otimes 1_5}$ (for $t\in [0,1]$ ). Then, for any $k>m$ and j ( $t\in [0,1]$ ),

(e7.9) $$ \begin{align} \gamma_k\left(h\circ \xi_j(t)\right)=\gamma_k\left(\gamma_m\left(f\circ \xi_i\circ \xi_j(t)\right)\otimes 1_5\right)=\gamma_m\left(f\circ\xi_i\circ \xi_j(t)\right)\otimes 1_{5R(k)}, \end{align} $$

where $R(k)$ is the rank of $\gamma _k$ and $\xi _i$ and $\xi _j$ are as defined in Lemma 6.8(1). Denote

(e7.10) $$ \begin{align} {\bar \gamma}_{k+1}(f)(t)=\gamma_{k+1}(f(t))\otimes 1_5\mathrm{\ for\ all\ } f\in C\left([0,1], M_{p_kq_k}(B)\right) \mathrm{\ (and\ } t\in [0,1]\text{)}. \end{align} $$

Therefore, from Lemma 6.8 and formula (e6.16) (also keep in mind Remark 6.10), we may write, for each $f\in A_m=E_{p_m,q_m}$ ,

(e7.11) $$ \begin{align} \varphi_{m, m+2}(f) =w_1^*\begin{pmatrix} H_0(f) &&& 0\\ &\bar\gamma_m\left(f\circ \xi^{(2)}_1\right)\otimes 1_{R(m+1)}&&\\ &&\ddots &\\ 0&&& \bar\gamma_m\left(f\circ \xi^{(2)}_{l(m+1)}\right)\otimes 1_{R(m+1)}\end{pmatrix} w_1, \end{align} $$

where $H_0: A_m\to C\!\left([0,1], M_{L_0p_mq_m}\right )$ is a homomorphism (for some integer $L_0\geq 1$ ), $w_1\in C\!\left([0,1], M_{p_{m+2}q_{m+2}}\right )$ is a unitary, $R(m+1)$ is the rank of $\bar \gamma _{m+1}$ and $\left \{\xi ^{(2)}_j: 1\le j\le l(m+1)\right \}$ is a full collection of compositions of two $\xi _i$ s (maps in Lemma 6.8(1)).

Therefore, by induction, for any $n>n(m)+m$ one may write, from the construction of Lemma 6.8 (see equation (e6.16)), for all $f\in A_m=E_{p_m,q_m}$ ,

(e7.12) $$ \begin{align} \varphi_{m,n}(f)= w^*\begin{pmatrix} H(f) &&&0 \\ & \bar\gamma_m(f\circ \Xi_1)\otimes 1_{R\left(n,1\right)} &&\\ &&\ddots &\\ 0&&& \bar\gamma_m(f\circ \Xi_l)\otimes 1_{R\left(n,l\right)}\end{pmatrix} w, \end{align} $$

where $H: A_m\to C\!\left([0,1], M_{Lp_mq_m}(B)\right )$ is a homomorphism (for some integer $L\ge 1$ ), $\Xi _j$ is a composition of $n-m$ maps in Lemma 6.8(1) such that the collection $\left \{\Xi _j: 1\le j\le l\right \}$ is full, $R(n,j)\ge 1$ is an integer, $j=1,2,\dotsc ,l$ , and $w\in C\!\left([0,1], M_{p_nq_n}\right )$ is a unitary.

It follows from Lemma 7.2 that

(e7.13) $$ \begin{align} \lvert\Xi_i(x)-\Xi_i(y)\rvert<(2/3)^{m-n}\mathrm{\ for\ all\ } x, y\in [0,1],\quad 1\le i\le l,\qquad\text{and}\qquad \cup_i^l \Xi_i([0,1])=[0,1]. \end{align} $$

Fix any $t\in [0,1]$ and $x\in [0,1]$ ; by Lemma 7.2, there are $y\in [0,1]$ and $j\in \{1,2,\dotsc ,l\}$ such that $\Xi _j(y)=x$ . Then

(e7.14) $$ \begin{align} \left\lvert\Xi_j(t)-x\right\rvert=\left\lvert\Xi_j(t)-\Xi_j(y)\right\rvert<(2/3)^{n-m}<(2/3)^{n(m)}. \end{align} $$

It follows from the choice of $n(m)$ and formula (e7.8) that, for $f\in {\cal F}_m$ ,

(e7.15) $$ \begin{align} \left\lVert\gamma_m\left(f\circ \Xi_j(t)\right)\right\rVert\ge (1-1/m)\lVert f\rVert\ge \left(\frac{63}{64}\right)^2 \mathrm{\ for\ all\ } t\in [0,1]. \end{align} $$

Since $\left \lVert f\left (\Xi _j(t)\right )-g\left (\Xi _j(t)\right )\right \rVert <1/64$ , this implies that

(e7.16) $$ \begin{align} \left\lVert\gamma_m\left(g\left(\Xi_j(t)\right)\right)\right\rVert\ge \frac{63^2-64}{64^2}\mathrm{\ for\ all\ } t\in [0,1]. \end{align} $$

Since for each $t\in [0,1]$ , we have $\gamma _m(g\circ \Xi _i(t))\in M_{p_mq_m}$ , $i=1,2,\dotsc ,l$ , we know that $ \varphi _{m,n}(g)(t) $ is not in any closed ideal of $M_{p_nq_n}(B)$ for each $t\in [0,1]$ . Therefore $\varphi _{m,n}(g)$ is full in $E_{p_n,q_n}=A_n$ . This proves the claim.

It follows from the claim that $A_z^C$ is simple. To see this, let $I\subset A_z^C$ be an ideal such that $I\not =A_z^C$ and put $C_n=\varphi _{n, \infty }(A_n)$ . Then $C_n\subset C_{n+1}$ for all n. Set $a\in C_m\setminus \{0\}$ . By the claim, there is $n'>m$ such that a is full in $C_{n'}$ , and therefore a is full in every $C_n$ for $n\ge n'$ . In other words, $a\not \in C_n\cap I$ for all n. It follows that $C_m\cap I=\{0\}$ , as $C_m\subset C_n$ for all $n\ge m$ . It is then standard to show that $I=\{0\}$ . Thus $A_z^C$ is simple.

Since, by Lemma 6.5, we have for each m that

$$ \begin{align*} \left(K_0(A_m), K_0(A_m)_+, \left[1_{A_m}\right], K_1(A_m)\right)=(\mathbb{Z}, \mathbb{Z}_+, 1, \{0\}), \end{align*} $$

one concludes (as each $\varphi _n$ is unital) that

(e7.17) $$ \begin{align} \left(K_0\left(A_z^C\right), K_0\left(A_z^C\right)_+, \left[1_{A_z^C}\right], K_1\left(A_z^C\right)\right)=(\mathbb{Z}, \mathbb{Z}_+, 1, \{0\}). \end{align} $$

Finally, if C is not exact, then B is not exact, since B has quotients of the form $\mathbb {C}\oplus C$ , which is not exact.

Define $\Phi : B\to C([0,1], M_{15}(B))$ by

(e7.18) $$ \begin{align} \Phi(f)(t):= \theta_t(f)\otimes 1_{15} \mathrm{\ for\ all\ } f\in B \mathrm{\ and\ } t\in [0,1], \end{align} $$

where $\theta _t: B\to B$ is defined in formula (e6.14). Note that for $f\in B$ ,

(e7.19) $$ \begin{align} \Phi(f)(0)=\theta_0(f)\otimes 1_{15}=f\otimes 1_{15}\in M_3(B)\otimes 1_5\qquad\text{and}\qquad \Phi(f)(1)=f(0)\otimes 1_{15}\in \mathbb{C}\cdot 1_{15}. \end{align} $$

One then obtains a unitary $u\in C([0,1], M_{15})$ such that

(e7.20) $$ \begin{align} u^*\Phi(f)u\in E_{3,5}. \end{align} $$

Define $\Psi (f):= u^*\Phi (f)u$ for all $f\in B$ . Then $\Psi $ is a unital injective homomorphism. In other words, B is embedded unitally into $A_1=E_{3,5}$ . Since each $\varphi _m: A_m\to A_{m+1}$ is unital and injective, B is embedded into $A_z^C$ . Since B is not exact, neither is $A_z^C$ (see, for example, [47, Proposition 2.6]).

Proposition 7.4. If C is exact but not nuclear, then $A_z^C$ is exact and not nuclear.

Proof. Note that since C is nonnuclear and exact, so is B. Note also that $A_n=E_{p_n,q_n}$ is a $C^*$ -subalgebra of the exact $C^*$ -algebra $C\!\left([0,1], M_{p_nq_n}(B)\right )$ . So each $A_n$ is exact. By [47, 2.5.5], $A_z^C$ is exact.

Let $\Phi : B\to A_1=E_{3,5}$ be as in the end of the proof of Theorem 7.3. Let $\pi _0^{(1)}: A_1\to M_3(B)\otimes M_5$ be the evaluation at $0$ , namely $\pi _0^{(1)}(f)=f(0)$ for all $f\in A_1$ . Let $\eta _1: M_3(B)\otimes \mathrm {1}_5\to B$ be given by defining $\eta _1\left (\left (b_{i,j}\right )_{ 3\times 3}\otimes 1_5 \right )=b_{1,1}$ , where $b_{i,j}\in B$ , $1\le i,j\le 3$ . Then $\eta _1$ is a norm $1$ c.p.c. map. Define $\pi _0^{\left (1,1\right )}: A_1\to B$ by $\pi _0^{\left (1,1\right )}(f):= \eta _1\circ \pi _0^{(1)}$ . Note that $\pi _0^{\left (1,1\right )}\circ \Phi $ is an isomorphism. In fact, $\pi _0^{\left (1,1\right )}\circ \Phi (b)=\theta _0(b)=b$ (see equation (e7.19)) for all $b\in B$ .

The foregoing is illustrated in the following diagram:

(e7.21)

We will use the same diagram in the n-stage.

In Lemma 6.8(4), let us denote $\xi _1$ such that $\xi _1(t)=2t/3$ for $t\in [0,3/4]$ and $\theta ^{(1)}(f(0))=f(0)$ for all $f\in E_{3,5}$ (note that we do not change the connecting map, but only for convenience in equation (e7.22)). So by formulas (e6.16), (e6.20), (e6.52) and (e6.53), we may write

(e7.22) $$ \begin{align} \varphi_1(f) \left(=\varphi_{1,2}(f)\right) =u_1^* \mathrm{diag}\left(\theta^{\left(1,2\right)}(f), H_1'(f)\right)u_1\mathrm{\ for\ all\ } f\in A_1, \end{align} $$

where $\theta ^{\left (1,2\right )}:=\theta ^{(1)'}: A_1\to C([0,1], M_{15}(B))$ and $\theta ^{\left (1,2\right )}(f)(0)=f(0)$ for $f\in A_1$ , and $H_1': A_1=E_{3,5}\to C\!\left([0,1], M_{p_1q_1}(B)\right )$ is a homomorphism. Note that the image of $H_1'$ is in a corner of $C\!\left([0,1], M_{p_1q_1}(B)\right )$ , and $u_1\in U\left (C\!\left([0,1], M_{p_2q_2}\right )\right )$ . Similarly, again by formulas (e6.16), (e6.20), (e6.52) and (e6.53), we may also write

(e7.23) $$ \begin{align} \varphi_{1,3} (f)=u_2^*\mathrm{diag}\left(\theta^{\left(1,3 \right)}(f), H_2'(f)\right)u_2\mathrm{\ for\ all\ } f\in A_1, \end{align} $$

where $\theta ^{\left (1,3 \right )}(f)(0)=f(0)$ for $f\in A_1$ , $H_2': A_1\to C\!\left([0,1], M_{p_2q_2}(B)\right )$ is a homomorphism and $u_2\in U\left (C\!\left([0,1], M_{p_3q_3}\right )\right )$ . By induction, for any $n>1$ we may write

(e7.24) $$ \begin{align} \varphi_{1,n}(f)=u_n^*\mathrm{diag}\left(\theta^{\left(1,n\right)}(f), H^{\prime}_n(f)\right)u_n\mathrm{\ for\ all\ } f\in A_1, \end{align} $$

where $\theta ^{\left (1,n\right )}(f)(0)=f(0)$ , $H^{\prime }_n: A_1\to C\!\left([0,1], M_{p_nq_n}(B)\right )$ is a homomorphism and $u_n\in C\!\left([0,1], M_{p_{n+1}q_{n+1}}\right )$ . (One should be warned that $u_n^*\mathrm {diag}\left (\theta ^{\left (1,n\right )},0,\dotsc ,0\right )u_n$ is not in $A_n$ .)

Now we prove that $A_z^C$ is not nuclear. We follow the proof of [13, Proposition 6]. Assume otherwise: For any finite subset ${\cal F}\subset B$ and $\varepsilon>0$ , if $A_z^C$ were nuclear, then $\varphi _{1, \infty }\circ \Phi $ would be nuclear. Therefore there would be a finite-dimensional $C^*$ -algebra D and c.p.c. maps ${\alpha }: B\to D$ and ${\beta }: D\to A_z^C$ such that

(e7.25) $$ \begin{align} \left\lVert\varphi_{1,\infty}\circ \Phi(b)-{\beta}\circ {\alpha}(b)\right\rVert<\varepsilon/2\mathrm{\ for\ all\ } b\in {\cal F}. \end{align} $$

Since $A_z^C$ is assumed to be nuclear, by the Effros–Choi lifting theorem [11], there exist an integer $n\ge 1$ and a unital c.p.c. map ${\beta }_n: D\to A_n$ such that

(e7.26) $$ \begin{align} \left\lVert{\beta}(x)-\varphi_{n, \infty}\circ {\beta}_n(x)\right\rVert<\varepsilon/2\mathrm{\ for\ all\ } x\in {\alpha}({\cal F}). \end{align} $$

Thus

(e7.27) $$ \begin{align} \left\lVert\varphi_{n, \infty}\left(\varphi_{1,n}\circ \Phi(b)-{\beta}_n\circ {\alpha}(b)\right)\right\rVert<\varepsilon. \end{align} $$

As $\varphi _{n,\infty }$ is an isometry, this implies that

(e7.28) $$ \begin{align} \left\lVert\varphi_{1,n}\circ \Phi(b)-{\beta}_n\circ {\alpha}(b)\right\rVert<\varepsilon\mathrm{\ for\ all\ } b\in B. \end{align} $$

Let $\pi _0^{(n)}: E_{p_n,q_n}\to M_{p_n}(B)\otimes 1_{q_n}$ be the evaluation at $0$ defined by $\pi _0^{(n)}(a):= a(0)$ . We have, by equation (e7.24),

(e7.29) $$ \begin{align} \pi_0^{(n)}\left(\varphi_{1,n}\circ \Phi(b)\right)=\mathrm{diag}\left(\theta_0(b) \otimes 1_{15}, H_n'(\Phi(f))(0)\right)\mathrm{\ for\ all\ } b\in B. \end{align} $$

Recall that $\theta _0(b)=b$ . Now a rank one projection p corresponding the first $(1,1)$ corner is in $M_{p_n}(B)\otimes 1_{q_n}$ . Put $q=u_n(0)^*pu_n(0)$ . We now use the n-stage diagram (e7.21). Let $\eta _n: M_{p_n}(B)\otimes 1_{q_n}\to B$ be defined by $\eta _n(x):= u_n(0)qxqu_n(0)^*$ for all $x\in M_{p_n}(B)\otimes 1_{q_n}$ which is a unital c.p.c. map $\eta _n\left (u_n(0)^*\left (\left (b_{i,j}\right )_{p_n\times p_n}\otimes 1_{q_n}\right )u_n(0)\right )=b_{1,1}$ . Note that $\eta _n\circ \pi _0^{(n)}\circ \varphi _{1,n}\circ \Phi =\mathrm {id}_B$ . By formula (e7.28),

(e7.30) $$ \begin{align} \left\lVert b-\eta_n\circ \pi_0^{(n)}\circ {\beta}_n\circ {\alpha}(b)\right\rVert=\left\lVert\eta_n\circ \pi_0^{(n)}\left(\varphi_{1,n}\circ \Phi(b)-{\beta}_n\circ {\alpha}(b)\right)\right\rVert<\varepsilon\mathrm{\ for\ all\ } b\in B. \end{align} $$

This would imply that B is nuclear. Therefore $A_z^C$ is not nuclear. The foregoing could be illustrated by the following diagram, which is only approximately commutative below the top triangle:

Theorem 7.5. The inductive limit $A_z^C$ in Theorem 7.3 has a unique tracial state.

Proof. First we note each unital $C^*$ -algebra $A_m=E_{p_m, q_m}$ has at least one tracial state, say $\tau _m$ . Note that $\varphi _{m, \infty }$ is an injective homomorphism. So we may view $\tau _m$ as a tracial state of $\varphi _{m, \infty }(A_m)$ . Extend $\tau _m$ to a state $t_m$ on $A_z^C$ . Choose a weak*-limit of $\{t_m\}$ , say t. Then t is a state of the unital $C^*$ -algebra $A_z^C$ . Note that $\varphi _{m, \infty }(A_m)\subset \varphi _{n, \infty }(A_n)$ if $n>m$ . Then for each pair $x, y\in \varphi _{m,\infty }(A_m)$ , and for any $n>m$ , $t_n(xy)=t_n(yx)$ . It follows that t is a tracial state of $A_z^C$ . In other words, $A_z^C$ has at least one tracial state.

Claim: For each k, each $a\in A_k$ with $\lVert a\rVert \le 1$ and each $\varepsilon>0$ , there exists $N>k$ such that, for all $n\ge N$ ,

(e7.31) $$ \begin{align} \left\lvert\tau_1\left(\varphi_{k, n}(a)\right)-\tau_2\left(\varphi_{k,n}(a)\right)\right\rvert<\varepsilon\mathrm{\ for\ all\ } \tau_1, \tau_2\in T(A_n). \end{align} $$

Fix $a\in A_k$ . To simplify the notation, without loss of generality we may assume that $k=1$ .

Choose $m>1$ such that

(e7.32) $$ \begin{align} 1/3^{m-1}<\varepsilon/4. \end{align} $$

Put $g=\varphi _{1,m}(a)$ . There is $\delta>0$ such that

(e7.33) $$ \begin{align} \lVert g(x)-g(y)\rVert<\varepsilon/4\mathrm{\ for\ all\ } x,y\in [0,1] \mathrm{\ with\ } \lvert x-y\rvert<\delta. \end{align} $$

Recall that here we view $\gamma _m$ as a map from $M_{p_mq_m}(B)$ to $M_{R(m)p_mq_m}$ . Note that for each $f\in A_m$ , since $\gamma _m(f(t))$ is a scalar matrix for all $t\in [0,1]$ , we have that $\gamma _m(f(t))(x)$ , as an element in $M_{R(m)p_mq_m}(B)$ , is a constant matrix (for $x\in (0,1]$ ) in $M_{R(m)p_mq_m}( C_0( (0,1], C)^{\sim })$ . Hence (see equation (e7.10) for $\bar \gamma _m$ ), for $t\in [3/4,1]$ – recalling that $\xi _i(t)=\xi _i(3/4)$ for all t in $[3/4,1]$ –

(e7.34) $$ \begin{align} \theta_{4\left(t-3/4\right)}\left(\bar\gamma_m\left(f\left(\Xi_j\circ \xi_i\right)\right)(3/4)\right)=\bar\gamma_m\left(f\left(\Xi_j\circ \xi_i\right)(3/4)\right)=\bar\gamma_m\left(f\left(\Xi_j\circ \xi_i\right)\right)(t). \end{align} $$

(Recall the definition of $\theta _t$ in formula (e6.21)). Therefore (see the definition of $\theta ^{(i)}$ in formula (e6.52)), for any i with $\xi _i(3/4)\not =1$ ,

(e7.35) $$ \begin{align} \theta^{(i)}\left(\bar\gamma_m\left(f\circ\Xi_j\right)\right)(t)&= \begin{cases} \bar\gamma_m\left(f\circ \Xi_j\circ \xi_i\right)(t) &\text{if } t\in [0,3/4],\\ \theta_{4\left(t-3/4\right)}\left(\bar\gamma_m\left(f\circ\Xi_j\circ \xi_i\right)(3/4)\right) &\text{if } t\in (3/4,1] \end{cases} \end{align} $$

(e7.36) $$ \begin{align} &= \bar\gamma_m\left(f\left(\Xi_j\circ \xi_i\right)\right)(t). \hspace{5.8pc} \end{align} $$

For those i such that $\xi _i(3/4)=1$ , one also has

(e7.37) $$ \begin{align} \theta^{(i)}\left(\bar\gamma_m\left(f\circ \Xi_j\right)\right)=\bar\gamma_m\left(f\circ\Xi_j\circ \xi_i\right). \end{align} $$

It follows (recall Lemma 6.8(4) for the definition of $\Theta _{m+1}$ , and also keep Remark 6.10 in mind) that

(e7.38) $$ \begin{align} \Theta_{m+1}(\varphi_m(f))=u^*\begin{pmatrix} \Theta_{m+1}'(f) & 0 &\dotsb & 0\\ 0 & \bar\gamma_{m}\left(f\circ \xi^{(2)}_1\right) &\dotsb &0\\ \vdots & \vdots & & \vdots\\ 0 & 0 & \dotsb & \bar\gamma_m\left(f\circ \xi^{(2)}_{k'}\right) \end{pmatrix} u \mathrm{\ for\ all\ } f\in A_m, \end{align} $$

where $u\in C\!\left([0,1], M_{5k_2p_{m+1}q_{m+1}}\right )$ is a unitary (the integer $k_2$ is the integer k in Lemma 6.8 for $\varphi _{m+1}$ ), $\Theta _{m+1}': A_m\to C\!\left([0,1], M_{T(0)p_mq_m}(B)\right )$ is a homomorphism for some integer $T(0)\ge 1$ and $\left \{\xi ^{(2)}_j: 1\le j\le k'\right \}$ is a full collection of compositions of two maps in Lemma 6.8(1). Moreover, by Lemma 6.8(2),

(e7.39) $$ \begin{align} T(0)/5k'R(m)<1/3^m. \end{align} $$

Then, combining with equation (e7.9), we may write $\varphi _{m,m+2}: A_m\to A_{m+2}$ as

(e7.40) $$ \begin{align} \varphi_{m,m+2}(f)=u_1^*\begin{pmatrix} H_{m+1}(f) & 0 &\dotsb & 0\\ 0 & \bar\gamma_{m}\left(f\circ \xi^{(2)}_1\right) \otimes 1_{r(1)} &\dotsb &0\\ \vdots & \vdots & & \vdots\\ 0 & 0 & \dotsb & \bar\gamma_m\left(f\circ \xi^{(2)}_{l(m)}\right)\otimes 1_{r(l(m))} \end{pmatrix} u_1 \end{align} $$

for all $f\in A_m$ , where $u_1\in C\!\left([0,1], M_{p_{m+2}q_{m+2}}\right )$ is a unitary, $H_{m+1}: A_m\to C\!\left([0,1], M_{T(1)p_mq_m}(B)\right )$ is a homomorphism for some integer $T(1)\ge 1$ , $\left \{\xi ^{(2)}_j: 1\le j\le l\right \}$ is a full collection of compositions of two maps in Lemma 6.8(1) and $r(l(j))\ge 1$ is an integer, $j=1,2,\dotsc ,l(m)$ . Moreover,

(e7.41) $$ \begin{align} T(1)/5R(m)\left(\sum_{j=1}^{l(m)} r(l(j))\right)<1/3^m. \end{align} $$

Therefore, by Lemma 6.8 – noting equations (e6.16), (e6.20) and (e7.36) – and the proof of Theorem 7.3 (see equation (e7.9)), as well as equation (e7.38) repeatedly, one may write, for each $n>m$ and all $f\in A_m$ ,

(e7.42) $$ \begin{align} \varphi_{m,n}(f)= w^*\begin{pmatrix} H_{m,n}(f) &&&0 \\ & \bar\gamma_m(f\circ \Xi_1) &&\\ &&\ddots &\\ 0&&& \bar\gamma_m(f\circ \Xi_L)\end{pmatrix} w, \end{align} $$

where $w\in C\!\left([0,1], M_{p_nq_n}\right )$ is a unitary, $H_{m,n}: A_m\to C\!\left([0,1], M_{L(0)p_mq_m}(B)\right )$ is a homomorphism for some integer $L(0)\ge 1$ , $\Xi _j:[0,1]\to [0,1]$ is a composition of $n-m$ many $\xi _i$ s and $\left \{\Xi _j:1\le j\le L\right \}$ is a full collection. Moreover,

(e7.43) $$ \begin{align} L(0)/5LR(m) <1/3^m. \end{align} $$

We choose N such that $(2/3)^{N-m}<\delta $ and choose any $n\ge N$ .

Set $\tau _i\in T(A_n)$ ( $i=1,2$ ). Then, for any $f\in A_n$ ,

(e7.44) $$ \begin{align} \tau_i(f)=\int_0^1 \sigma_i(t) (f(t))d\mu_i,\quad i=1,2, \end{align} $$

where $\sigma _i(t)$ is a tracial state of $M_{p_nq_n}(B)$ for all $t\in (0,1)$ , $\sigma _i(0)$ is a tracial state of $M_{p_n}(B)\otimes 1_{q_n}$ , $\sigma _i(1)$ is a tracial state of $1_{p_n}\otimes M_{q_n}$ and $\mu _i$ is a probability Borel measure on $[0,1]$ , $i=1,2$ . For each $t\in [0,1]$ and for $f(t)\in M_{p_nq_n}\subset M_{p_nq_n}(B)$ ,

(e7.45) $$ \begin{align} \sigma_i(t)(f(t))=\mathrm{tr}(f(t)),\quad i=1,2, \end{align} $$

where $\mathrm {tr}$ is the normalised trace on $M_{p_nq_n}$ (see equation (e6.11)). For each $j\in \{1,2,\dotsc ,L\}$ , by Lemma 7.2,

(e7.46) $$ \begin{align} \left\lvert\Xi_j(x)-\Xi_j(y)\right\rvert<(2/3)^{n-m}<\delta\mathrm{\ for\ all\ } x, y\in [0,1]. \end{align} $$

By the choice of $\delta $ ,

(e7.47) $$ \begin{align} \left\lVert g\circ \Xi_j(x) -g\circ \Xi_j(y)\right\rVert<\varepsilon/4\mathrm{\ for\ all\ } x, y\in [0,1]. \end{align} $$

For each $f\in A_m$ , write

(e7.48) $$ \begin{align} H'(f)(t)=\begin{pmatrix} H_{m+1}(f)(t) & 0\\ 0 & 0\end{pmatrix}\mathrm{\ for\ all\ } t\in [0,1]. \end{align} $$

Then one has, for each $f\in A_m$ and $i=1,2$ ,

(e7.49) $$ \begin{align} \tau_i\left(\varphi_{m,n}(f)\right)&=\int_0^1 \sigma_i(t)\left(\varphi_{m,n}(f)\right) d\mu_i \hspace{12pc} \end{align} $$

(e7.50) $$ \begin{align} & \hspace{0.5pc} =\int_0^1\sigma_i(t)(H'(f)(t))d\mu_i +\int_0^1 \mathrm{tr}\left(\bigoplus_{j=1}^L\left(f\circ \Xi_j(t)\right)\right)d\mu_i. \end{align} $$

By formula (e7.47), recalling that $\lVert g\rVert \le 1$ ,

(e7.51) $$ \begin{align} \int_0^1\left\lvert\mathrm{tr}\left(\bigoplus_{j=1}^L(g\circ \Xi_j(1/2)\right)-\bigoplus_{j=1}^L \left(g\circ \Xi_j(t)\right)\right\rvert d\mu_i <(\varepsilon/4)\int_0^1d\mu_i =\varepsilon/4. \end{align} $$

By formula (e7.43),

(e7.52) $$ \begin{align} \int_0^1\lvert\sigma_i(t)(H'(g)(t))\rvert d\mu_i <(1/3)^m<\varepsilon/4. \end{align} $$

Recall that $\varphi _{1,n}(a)=\varphi _{m, n}(g)$ . Thus, by formulas (e7.49), (e7.50), (e7.51) and (e7.52),

(e7.53) $$ \begin{align} \left\lvert\tau_i\left(\varphi_{1,n}(a)\right)-\sum_{j=1}^L\mathrm{tr}\left(g\left(\Xi_j(1/2)\right)\right)\right\rvert<\varepsilon/2,\quad i=1,2. \end{align} $$

Therefore,

(e7.54) $$ \begin{align} \left\lvert\tau_1\left(\varphi_{1,n}(a)\right)-\tau_2\left(\varphi_{1,n}(a)\right)\right\rvert<\varepsilon. \end{align} $$

This proves the claim.

To complete the proof, set $s_1, s_2\in T\left (A_z^C\right )$ . Set $a\in A_z^C$ and $\varepsilon>0$ . Then there is $f\in A_k$ for some $k\ge 1$ such that

(e7.55) $$ \begin{align} \left\lVert a-\varphi_{k,\infty}(f)\right\rVert<\varepsilon/3. \end{align} $$

Let $\tau _{i,n}=s_i\circ \varphi _{n,\infty }$ . Then, by the claim, there exists $N\ge k$ such that for all $n>N$ ,

(e7.56) $$ \begin{align} \left\lvert\tau_{1,n}\left(\varphi_{k,n}(f)\right)-\tau_{2 ,n}\left(\varphi_{k,n}(f)\right)\right\rvert<\varepsilon/3. \end{align} $$

It follows that

(e7.57) $$ \begin{align} \left\lvert s_1\left(\varphi_{k,\infty}(f)\right)-s_2\left(\varphi_{k,\infty}(f)\right)\right\rvert\le \varepsilon/3. \end{align} $$

Therefore,

$$ \begin{align*} \lvert s_1(a)-s_2(a)\rvert &\le \left\lvert s_1(a)-s_1\left(\varphi_{k,\infty}(f)\right)\right\rvert\\ &\quad +\left\lvert s_1\left(\varphi_{k,\infty}(f)\right)-s_2\left(\varphi_{k,\infty}(f)\right)\right\rvert+\left\lvert s_2(a)-s_2\left(\varphi_{k,\infty}(f)\right)\right\rvert <\varepsilon. \end{align*} $$

It follows that $s_1(a)=s_2(a)$ . Thus $A_z^C$ has a unique tracial state.

Remark 7.6. Recall that the construction allows $B=\mathbb {C}$ (with $C=\{0\}$ ). In that case, of course, $A_z^C={\mathscr Z}$ . Note that when $B=\mathbb {C}$ , we have $\theta _t(b)=b$ for all $b\in M_{p_mq_m}$ . In other words, $\theta _t=\operatorname {id}_B$ .

Let

(e7.58) $$ \begin{align} Z_{p_m,q_m}=\left\{f\in C\left([0,1], M_{p_mq_m}\right): f(0)\in M_{p_m}\otimes 1_{q_m}\mathrm{\ and\ } f(1)\in 1_{p_m}\otimes M_{q_m}\right\}. \end{align} $$

In general (when $C\not =\{0\}$ ), one has $Z_{p_m,q_m}\subset E_{p_m,q_m}$ , as we view $\mathbb {C}\subset B$ and $M_{p_mq_m}\subset M_{p_mq_m}(B)$ . Let $\varphi _m^z={\varphi _m}\rvert _{Z_{p_m,q_m}}$ . Then, since $v\in C\!\left([0,1], M_{p_{m+1}q_{m+1}}\right )$ (see the line before formula (e6.48)), $\varphi _m^z\left (Z_{p_m,q_m}\right )\subset Z_{p_{m+1}, q_{m+1}}$ . Thus, one obtains a unital $C^*$ -subalgebra (of $A_z^C$ )

(e7.59) $$ \begin{align} B_z=\lim_{n\to\infty}\left(Z_{p_m, q_m}, \varphi_m^z\right). \end{align} $$

Then $B_z\cong {\mathscr Z}$ [27].

8. Regularity properties of $A_z^C$

In this section, let $A_z^C$ be the $C^*$ -algebra in Theorem 7.3.

Lemma 8.1. The inductive system can be arranged so that $A_z^C$ has the following properties:

1. (1) $A_z^C$ has a unital $C^*$ -subalgebra $B_z\cong {\mathscr Z}$ .

2. (2) For any finite subset ${\cal F}\subset A_m$ and $\varepsilon>0$ , there is $e\in (A_{m+1})_+^{\textbf{1}}\setminus \{0\}$ such that the following hold:

   1. (i) $e(t)\in M_{p_{m+1}q_{m+1}}$ for all $t\in [0,1]$ and $e(1)=0$ .

   2. (ii) $\lVert ex-xe\rVert <\varepsilon \text { for all } x\in \varphi _m({\cal F})$ .

   3. (iii) $\varphi _{m+1, \infty }\left ((1-e)^{\beta }\varphi _m(f)\right )\in _{\varepsilon } B_z$ for all $f\in {\cal F}$ , and for any ${\beta } \in (0, \infty )$ ,

      (e8.1) $$ \begin{align} \left\lVert\varphi_{m+1, \infty}\left((1-e)^{{\beta}}\varphi_m(y)\right)\right\rVert\ge (1-\varepsilon)\lVert\varphi_m(y)\rVert \mathrm{\ for\ all\ } y\in {\cal F}_m. \end{align} $$

   4. (iv) $d_\tau (e)<1/3^m$ for all $\tau \in T(A_{m+1})$ .

(Recall that ${\cal F}_m$ was constructed in the proof of Theorem 7.3.)

Proof. We will keep the notation used in the proof of Lemma 6.8.

For (i), we note that the $C^*$ -subalgebra $B_z=\lim _{n\to \infty }\left (Z_{p_m,q_m}, {\varphi _m}\rvert _{Z_{p_m,q_m}}\right )$ has been identified in Remark 7.6, where

(e8.2) $$ \begin{align} Z_{p_m,q_m}=\left\{f\in C\left([0,1], M_{p_mq_m}\right): f(0)\in M_{p_m}\otimes 1_{q_m}\mathrm{\ and\ } f(1)\in 1_{p_m}\otimes M_{q_m}\right\}. \end{align} $$

There is $\delta \in (0,\varepsilon /2) $ such that if $\lvert t-t'\rvert <2\delta $ ,

(e8.3) $$ \begin{align} \lVert\varphi_m(f)(t)-\varphi_m(f)(t')\rVert <\varepsilon/4\mathrm{\ for\ all\ } f\in {\cal F}. \end{align} $$

In particular, there is $t_1\in (0,1)$ ( $1-t_1<\delta $ ) such that

(e8.4) $$ \begin{align} \lVert\varphi_m(f)(t)-\varphi_m(f)(1)\rVert<\varepsilon/4\mathrm{\ for\ all\ } f\in {\cal F} \mathrm{\ and\ } t\in (t_1,1). \end{align} $$

Choose a continuous function $g\in C([0,1])$ such that $0\le g\le 1$ , $g(t)=1$ for all $t\in [0,t_1]$ and $g(t)=(1-t)/(1-t_1)$ for $t\in (t_1,1]$ . Let $e_0(t)=g(t)\cdot 1_{A_m}$ for all $t\in [0,1]$ . Note that $e_0(0)=1_{p_mq_m}\in M_{p_m}(B)\otimes 1_{q_m}$ and $e_0(1)=0\in 1_{p_m}\otimes M_{q_m}$ . So $e_0\in A_m$ . Moreover, $e_0$ is in the center of $A_m$ . Define $\sigma _0: M_{p_m}(B)\otimes M_{q_m}\to M_{d_mp_m}(B)\otimes M_{5q_m}$ by $\sigma _0'\otimes s$ , where

(e8.5) $$ \begin{align} \sigma_0'(a)=\begin{pmatrix}\theta_1(a) &0\\ 0 & 0\end{pmatrix}\mathrm{\ for\ all\ } a\in M_{p_m}(B)\mathrm{\ and\ } s(c)=c\otimes 1_5 \mathrm{\ for\ all\ } c\in M_{q_m}, \end{align} $$

where $\theta _1: M_{p_m}(B)\to M_{p_m}\subset M_{p_m}(B)$ is defined by $\theta _1(c)(x) := c(0)$ for $c\in M_{p_m}(B)=\, M_{p_m}( C_0( (0,1], C)^{\sim })$ and for all $x\in [0,1]$ , and the “ $0$ ” in the lower corner of the matrix has the size of $(d_m-1)p_m\times (d_m-1)p_m$ . Then define $\sigma _1: A_m\to C\!\left([0,1], M_{d_mp_m5q_m}(B)\right )$ by

(e8.6) $$ \begin{align} \sigma_1(f)(t):= \sigma_0(f(t))\mathrm{\ for\ all\ } f\in E_{p_m,q_m} \mathrm{\ and\ } t\in [0,1]. \end{align} $$

It follows that for all fixed $t\in [0,1]$ ,

(e8.7) $$ \begin{align} \sigma_1(e_0)(t)&=\sigma_0(e_0(t)) =\sigma_0\left(g(t)\cdot 1_{A_m}\right) =\sigma_0\left(g(t)\cdot 1_{p_m}\otimes 1_{q_m}\right) \hspace{9pc} \end{align} $$

(e8.8) $$ \begin{align} & \hspace{3pc} =\begin{pmatrix} \left(\theta_1(g(t)\cdot 1_{p_m}\right)&0\\ 0 & 0\end{pmatrix}\otimes 1_{5q_m} =\begin{pmatrix} g(t)\cdot 1_{p_m}&0\\ 0 & 0\end{pmatrix}\otimes 1_{5q_m}=\begin{pmatrix} g(t)\cdot 1_{p_mq_m} &0\\ 0 & 0\end{pmatrix}\otimes 1_{5}, \end{align} $$

where the last “0” in the last matrix has the size $(d_m-1)p_mq_m\times (d_m-1)p_mq_m$ . Thus

(e8.9) $$ \begin{align} \sigma_1(e_0)(0) =b\otimes 1_{5q_m} \qquad\text{and}\qquad \sigma_1(e_0(1))=0, \end{align} $$

where $ b=\begin {pmatrix} 1_{p_m} &0\\ 0 & 0\end {pmatrix}$ . It follows that $\sigma _1(e_0)\in E_{d_mp_m, 5q_m}$ . Note that for each $\tau \in T(A_m)$ , by formula (e 6.13).

(e8.10) $$ \begin{align} d_\tau(\sigma_1(e_0))<1/3^m. \end{align} $$

Let us recall the definition of $\tilde \psi _{m,i}$ in the proof of Lemma 6.8, $1\le i\le k$ (see formula (e6.50)). Then for all $f\in A_m$ , by formulas (e6.50), (e6.44), (e6.46) and (e6.52), for each $t\in [0,1]$ we have

(e8.11) $$ \begin{align} \tilde\psi_{m,i}(f)(t)\sigma_1(e_0)(t) &=\begin{pmatrix} \theta^{(i)}(f)(t) & 0\\ 0 & \gamma_m(f(t))\end{pmatrix} \otimes 1_5 \cdot \begin{pmatrix} g(t)\cdot 1_{p_mq_m} &0\\ 0 & 0\end{pmatrix}\otimes 1_{5} \end{align} $$

(e8.12) $$ \begin{align} &=\begin{pmatrix} \theta^{(i)}(f)(t)\cdot g(t)\cdot 1_{p_mq_m}& 0\\ 0 & 0\end{pmatrix}\otimes 1_5 \hspace{1.5pc} \end{align} $$

(e8.13) $$ \begin{align} & \hspace{4.5pc} =\begin{pmatrix} g(t)\cdot 1_{p_mq_m} & 0\\ 0 & 0\end{pmatrix} \otimes 1_5 \cdot \begin{pmatrix} \theta^{(i)}(f)(t) &0\\ 0 & \gamma_m(f(t))\end{pmatrix}\otimes 1_{5} \end{align} $$

(e8.14) $$ \begin{align} &= \sigma_1(e_0)(t){\tilde \psi}_{m,i}(f)(t). \hspace{5.5pc} \end{align} $$

In other words, for all $f\in E_{p_m,q_m}$ ,

(e8.15) $$ \begin{align} \tilde\psi_{m,i}(f)\sigma_1(e_0)=\sigma_1(e_0)\tilde\psi_{m,i}(f),\quad i=1,2,\dotsc,k. \end{align} $$

Define ${\alpha }: [0,1]\to [0,1]$ by

(e8.16) $$ \begin{align} {\alpha}(t):= \begin{cases} {\frac{t}{t_1}}& \text{if } t\in [0,t_1],\\ 1& \text{if } t\in (t_1,1]. \end{cases} \end{align} $$

Then for all j, $f\circ {\alpha }\in E_{p_j,q_j}$ if $f\in E_{p_j,q_j}$ . Moreover, by formula (e8.3),

(e8.17) $$ \begin{align} \lVert\varphi_m(f)\circ {\alpha} -\varphi_m(f)\rVert<\varepsilon/4\mathrm{\ for\ all\ } f\in {\cal F}. \end{align} $$

Therefore, for each $f\in A_m$ , each $t\in [0,1]$ and each ${\beta }\in (0,\infty )$ , with $l=d_mp_m5q_m$ ,

(e8.18) $$ \begin{align} (1_l-\sigma_1(e_0))^{\beta}{\tilde\psi_{m,i}}(f)\circ {\alpha} (t)=\begin{pmatrix} (1-g(t))^{\beta}\cdot 1_{p_mq_m}\cdot \theta^{(i)}(f)\circ {\alpha}(t) & 0\\ 0& \gamma_m(f(t))\end{pmatrix} \otimes 1_5, \end{align} $$

for $i=1,2,\dotsc ,k$ . For $t\in [0,t_1]$ , by the definition of g,

(e8.19) $$ \begin{align} (1-g(t))^{\beta}\cdot 1_{p_mq_m}\cdot \theta^{(i)}(f)(t)=0. \end{align} $$

For $t\in (t_1,1]$ ,

(e8.20) $$ \begin{align} (1-g(t))^{\beta}\cdot 1_{p_mq_m}\cdot \theta^{(i)}(f)\circ {\alpha} (t)=(1-g(t))^{\beta}\cdot \theta^{(i)}(f)(1)\in M_{p_mq_m}. \end{align} $$

Hence

(e8.21) $$ \begin{align} (1_l-\sigma_1(e_0))^{\beta}\tilde\psi_{m,i}(f\circ {\alpha}) \in C\left([0,1], M_{d_mp_m5q_m}\right). \end{align} $$

Moreover, by formula (e7.8), for $f\in {\cal F}_m$ we have

(e8.22) $$ \begin{align} \left\lVert(1_l-\sigma_1(e_0))^{\beta}\tilde\psi_{m,i}(f)\circ {\alpha} \right\rVert\ge (1-1/m)\lVert f\rVert. \end{align} $$

Using the same v as in formula (e6.48), with $\sigma _1(e_0)(t)$ repeating k times, define

(e8.23) $$ \begin{align} e:=v(t)^*\begin{pmatrix} \sigma_1(e_0)(t) & 0 &\dotsb & 0\\ 0 & \sigma_1(e_0)(t) &\dotsb &0\\ \vdots & \vdots & & \vdots\\ 0 & 0 & \dotsb & \sigma_1(e_0)(t)\end{pmatrix}v(t). \end{align} $$

With b as in the line after equation (e8.9), $b\otimes 1_{5q_m}\otimes 1_{r_0}=b\otimes 1_{r_05q_m}=\left (b\otimes 1_{t_0}\right )\otimes 1_{q_{m+1}}$ (see formula (e6.35)). Since $\sigma _1(e_0)\in E_{d_mp_m,5q_m}$ , as in equation (e8.9),

(e8.24) $$ \begin{align} e(0)=v_0^*\begin{pmatrix} b\otimes 1_{5q_m} & 0 &\dotsb & 0\\ 0 & b\otimes 1_{5q_m} &\dotsb &0\\ \vdots & \vdots & & \vdots\\ 0 & 0 & \dotsb & b\otimes 1_{5q_m}\end{pmatrix}v_0\in M_{p_{m+1}}\otimes 1_{q_{m+1}} \end{align} $$

(see formula (e6.37)). Combining with the fact that $e(1)=0$ , one concludes that $e\in E_{p_{m+1},q_{m+1}}=A_{m+1}$ . Moreover, by equation (e8.7) and the fact that $v\in C\!\left([0,1], M_{p_{m+1}q_{m+1}}\right )$ (see discussion before formula (e6.48)), $e(t)\in M_{p_{m+1}q_{m+1}}$ for each $t\in [0,1]$ and $e(1)=0$ . So part (2)(i) of the statement of the lemma holds.

By equation (e8.15) and (e6.51), one computes that for all $f\in A_m$ ,

(e8.25) $$ \begin{align} e\varphi_m(f)&=v^*\begin{pmatrix} \sigma_1(e_0)\tilde\psi_{m,1}(f) & &&0 \\ & & \ddots& \\ 0&& & \sigma_1(e_0)\tilde\psi_{m,k}(f)\end{pmatrix}v \hspace{5.5pc} \end{align} $$

(e8.26) $$ \begin{align} & \hspace{1pc} =v^*\begin{pmatrix} \tilde\psi_{m,1}(f)\sigma_1(e_0) & &&0 \\ & & \ddots& \\ 0&& & \tilde\psi_{m,k}(f)\sigma_1(e_0)\end{pmatrix}v =\varphi_m(f) e. \end{align} $$

In other words, part (2)(ii) in the statement of the lemma holds. Now

$$ \begin{align*} (1-e)^{\beta}\varphi_m(f\circ {\alpha}) =v^*\begin{pmatrix} (1_l-\sigma_1(e_0))^{\beta}\tilde\psi_{m,1}(f)\circ {\alpha} & &&0 \\ & & \ddots& \\ 0&& & (1_l-\sigma_1(e_0))^{\beta} \tilde\psi_{m,k}(f)\circ {\alpha} \end{pmatrix}v \end{align*} $$

for all $f\in A_m$ , where $l=d_mp_m5q_m$ . Note that $(1-e)^{\beta }\varphi _m(f)\circ {\alpha } \in E_{p_{m+1}, q_{m+1}}$ . It follows from formula (e8.21) that

(e8.27) $$ \begin{align} (1-e)^{\beta}\varphi_m(f)\circ {\alpha} \in Z_{p_{m+1}, q_{m+1}}\mathrm{\ for\ all\ } f\in {\cal F}. \end{align} $$

Then, by formula (e8.17),

(e8.28) $$ \begin{align} (1-e)^{\beta}\varphi_m(f)\in_{\varepsilon/4} Z_{p_{m+1}, q_{m+1}}\mathrm{\ for\ all\ } f\in {\cal F}. \end{align} $$

It follows that

(e8.29) $$ \begin{align} \varphi_{m, \infty}\left((1-e)^{\beta} f\right)\in_{\varepsilon} B_z\mathrm{\ for\ all\ } f\in {\cal F}. \end{align} $$

Moreover, by formula (e8.22), formula (e8.1) also holds, so part (2)(iii) of the statement of the lemma holds. It follows from formula (e8.10) that part (2)(iv) also holds.

Lemma 8.2. Let

(e8.30) $$ \begin{align} E_{p,q}=\left\{(f,c): C\left([0,1], M_{pq}(B)\right)\oplus \left(M_p(B)\oplus M_q\right): \pi_0(c)=f(0)\mathrm{\ and\ } \pi_1(c)=f(1)\right\}, \end{align} $$

where $\pi _0: M_p(B)\oplus M_q \to M_p(B)\otimes 1_q\subset M_{pq}(B)$ defined by $\pi _0(c_1\oplus c_2):= c_1\otimes 1_q$ for all $c_1\in M_p(B)$ and $c_2\in M_q$ , and $\pi _1: M_p(B)\oplus M_q \to 1_p\otimes M_q\subset M_{pq}(B)$ defined by $\pi _1((c_1\oplus c_2)):= 1_p\otimes c_2$ for all $c_1\in M_p(B)$ and $c_2\in M_q$ (see formula (e6.6)). Let

(e8.31) $$ \begin{align} L_{p,q}=\left\{(f,c): C\left([0,1], M_{pq}\right)\oplus M_p: \pi_0\rvert_{M_p}(c)=f(0)\right\}, \end{align} $$

where $\pi _0\rvert _{M_p}(c) =c\otimes 1_q$ for all $c\in M_p$ .

Suppose that $a, b\in {E_{p,q}}_+$ are such that

1. (1) $a(t)\in C\!\left([0,1], M_{pq}\right )$ and $a(1)=0$ and

2. (2) there is $b_0\in C\!\left([0,1], M_{pq}\right )_+$ such that $b_0(t)\le b(t)$ for all $t\in [0,1]$ and

   (e8.32) $$ \begin{align} a\lesssim b_0\mathrm{\ in\ } L_{p,q} \end{align} $$
   (i.e., there exists a sequence $x_n\in L_{p,q}$ such that $x_n^*b_0x_n\to a$ ).

Then

(e8.33) $$ \begin{align} a\lesssim b\mathrm{\ in\ } E_{p,q}. \end{align} $$

Proof. Let $1>\varepsilon >0$ . Consider a continuous function $h_\delta \in E_{p,q}$ :

(e8.34) $$ \begin{align} h_\delta(t)=\begin{cases} 1_{M_{pq}(B)} &\text{if } t\in [0,1-\delta],\\ 0&\text{if } t\in (1-\delta/2,1],\\ \text{linear} & \text{otherwise.} \end{cases} \end{align} $$

Since $a(1)=0$ , there exists $\delta _0>0$ such that $\left \lVert a-h_{\delta _0}^{1/2}a\cdot h_{\delta _0}^{1/2}\right \rVert <\varepsilon $ .

Note that $h_{\delta _0}$ lies in the center of $C\!\left([0,1], M_{pq}(B)\right )$ , and for any $f\in L_{p,q}$ and any $n\in \mathbb {N}$ , we have $h^{1/n}_{\delta _0}\cdot f\in E_{p,q}$ . Then since $a\lesssim b_0$ in $L_{p,q}$ , one checks $h_{\delta _0}^{1/2}a h_{\delta _0}^{1/2}\lesssim h_{\delta _0}^{1/2}b_0 h_{\delta _0}^{1/2}\le b_0\le b$ in $E_{p,q}$ . This implies that $a\approx _{\varepsilon } h_{\delta _0}^{1/2}a h_{\delta _0}^{1/2}\lesssim b$ in $E_{p,q}$ . Since this holds for any $1>\varepsilon >0$ , one concludes that $a\lesssim b$ in $E_{p,q}$ .

Definition 8.3 compare [19, 2.1.1.]

In the spirit of Definition 3.1, a simple $C^*$ -algebra A is said to have essential tracial nuclear dimension at most n if A is essentially tracially in ${\cal N}_n$ , the class of $C^*$ -algebras with nuclear dimension at most n – that is, if for any $\varepsilon>0$ and any finite subsets ${\cal F}\subset A$ and $a\in A_+\setminus \{0\}$ , there exist an element $e\in A_+^{\textbf{1}}$ and a $C^*$ -subalgebra $B\subset A$ which has nuclear dimension at most n such that

1. (1) $\lVert ex-xe\rVert <\varepsilon $ for all $x\in {\cal F}$ ,

2. (2) $(1-e)x\in _{\varepsilon } B$ and $\lVert (1-e)x\rVert \ge \lVert x\rVert -\varepsilon $ for all $x\in {\cal F}$ and

3. (3) $e\lesssim a$ .

Let us denote by ${\cal N}_{{\mathscr Z},s,s}$ the class of separable nuclear simple ${\mathscr Z}$ -stable $C^*$ -algebras.

Theorem 8.4. The unital simple $C^*$ -algebra $A_z^C$ is essentially tracially in ${\cal N}_{{\mathscr Z},s,s}$ and has essential tracial nuclear dimension at most $1$ , stable rank one and strict comparison for positive elements. Moreover, $A_z^C$ has a unique tracial state and has no $2$ -quasitraces other than the unique tracial state, and

(e8.35) $$ \begin{align} \left(K_0\left(A_z^C\right), K_0\left(A_z^C\right)_+, \left[1_{A_z^C}\right], K_1\left(A_z^C\right)\right)=(\mathbb{Z}, \mathbb{Z}_+, 1, \{0\}). \end{align} $$

Recall that if C is exact and not nuclear, then $A_z^C$ is exact and not nuclear (Theorem 7.3), and if $C$ is not exact, then $A_z^C$ is not exact (Proposition 7.4).

Proof. We will first show that $A_z^C$ is essentially tracially in ${\cal N}_{{\mathscr Z},s,s}$ . We will retain the notation from the construction of $A_z^C$ .

Fix a finite subset ${\cal F}$ and an element $a\in {A_z^C}_+$ with $\lVert a\rVert =1$ . To verify that $A_z^C$ has the specified property, without loss of generality we may assume that there is a finite subset ${\cal G}\subset A_1^{\textbf{1}}$ such that $\varphi _{1, \infty }({\cal G})={\cal F}$ . Since $\cup _{n=1} {\cal F}_{1,n}$ is dense in $A_1$ (see the proof of Theorem 7.3), we may assume that ${\cal G}\subset {\cal F}_{1,m}$ for some $m\ge 1$ . By the first few lines of the proof of Theorem 7.3, we may assume that $\varphi _{1,m}({\cal G})\subset \varphi _{1, m}\left ({\cal F}_{1,m}\right )\subset {\cal F}_{m+1,1}$ . Starting from $m+1$ instead of $1$ , to further simplify notation, without loss of generality we may assume that ${\cal G}\subset {\cal F}_{1,1}={\cal F}_1\cup \{0\}$ . Without loss of generality, we may assume that there is $a'\in (A_1)_+^{\textbf{1}}$ with $\lVert a'\rVert =1$ such that

(e8.36) $$ \begin{align} \lVert\varphi_{1, \infty}(a')-a\rVert<1/4. \end{align} $$

It follows from [40, Proposition 2.2] that

(e8.37) $$ \begin{align} \varphi_{1, \infty}\left(f_{1/4}(a')\right)=f_{1/4}\left(\varphi_{1, \infty}(a')\right)\lesssim a. \end{align} $$

Put $a_0'=f_{1/4}(a')\ (\neq 0)$ . Since $A_z^C$ is simple, there are $x_1,x_2,\dotsc ,x_k\in A_z^C$ such that

(e8.38) $$ \begin{align} \sum_{i=1}^k x_i^* \varphi_{1, \infty}\left(a_0'\right)x_i=1. \end{align} $$

It follows that for some large $n_0$ , there are $y_1,y_2,\dotsc ,y_k\in A_{n_0}$ and $n_1\ge n_0$ such that

(e8.39) $$ \begin{align} \left\lVert\sum_{i=1}^k \varphi_{n_0, n_1}\left(y_i^*\right) \varphi_{1, n_1}(a_0')\varphi_{n_0, n_1}(y_i)-1_{A_{n_1}}\right\rVert<1/4. \end{align} $$

It follows that $a_0:=\varphi _{1, n_1}\left (a_0'\right )$ is a full element in $A_{n_1}$ .

Set

(e8.40) $$ \begin{align} d=\inf\left\{d_\tau(a_0): \tau\in T\left(A_{n_1}\right)\right\}. \end{align} $$

Since $a_0$ is full in $A_{n_1}$ and $a_0\in \left (A_{n_1}\right )_+$ , we have $d>0$ . Choose $m>n_1$ such that

(e8.41) $$ \begin{align} d/4>1/3^{m-1}. \end{align} $$

By applying Lemma 8.1, we obtain $e\in (A_{m+1})_+^{\textbf{1}}\setminus \{0\}$ such that

1. (i) $e(t)\in M_{p_{m+1}q_{m+1}}$ for all $t\in [0,1]$ and $e(1)=0$ ,

2. (ii) $\lVert ex-xe\rVert <\varepsilon \text { for all } x\in \varphi _m\left (\varphi _{1, m}({\cal G})\right )$ ,

3. (iii) $\varphi _{m+1, \infty }\left ((1-e)\varphi _m\left (\varphi _{1, m}(x)\right )\right )\in _{\varepsilon } B_z$ , and

   $$ \begin{align*} \left\lVert\varphi_{m+1, \infty}\left((1-e)\varphi_m\left(\varphi_{1, m}(x)\right)\right)\right\rVert\ge (1-1/m)\left\lVert\varphi_{1,m}(x)\right\rVert \end{align*} $$
   for all $x\in {\cal F}_{1,1}$ , and

4. (iv) $d_\tau (e)<1/3^m$ for all $\tau \in T(A_{m+1})$ .

Denote $a_1=\varphi _{n_1,m}(a_0)$ . It is full in $A_m$ . Write, as in Theorem 6.8 and equation (e6.16),

(e8.42) $$ \begin{align} \varphi_{m}(a_1)=u^*\begin{pmatrix} \Theta_{m}(a_1) & 0 &\dotsb & 0\\ 0 & \gamma_{m}(a_1\circ \xi_1)\otimes 1_5 &\dotsb &0\\ \vdots & \vdots & & \vdots\\ 0 & 0 & \dotsb & \gamma_m(a_1\circ \xi_k)\otimes 1_5 \end{pmatrix} u, \end{align} $$

where $u\in U\left (C\!\left([0,1], M_{p_{m+1}q_{m+1}}(B)\right )\right )$ , $\Theta _{m}: A_m\to C\!\left([0,1], M_{kp_mq_m}(B)\right )$ is a homomorphism, $k\ge 1$ is an integer and $\gamma _m: B\to M_{R(m)}$ is a finite-dimensional representation. Moreover,

(e8.43) $$ \begin{align} k/5kR(m)<1/3^{m}. \end{align} $$

Let

(e8.44) $$ \begin{align} b_0=u^*\begin{pmatrix} 0 & 0 &\dotsb & 0\\ 0 & \gamma_{m}(a_1\circ \xi_1)\otimes 1_5 &\dotsb &0\\ \vdots & \vdots & & \vdots\\ 0 & 0 & \dotsb & \gamma_m(a_1\circ \xi_k)\otimes 1_5 \end{pmatrix} u. \end{align} $$

Note that $b_0\in C\!\left([0,1], M_{p_{m+1}q_{m+1}}\right )$ . Moreover, since $a_1\in E_{p_m,q_m}$ , we have $a_1(0)=a_1'\otimes 1_{q_m}$ for some $a_1'\in M_{p_m}(B)$ . Therefore,

(e8.45) $$ \begin{align} \gamma_m(a_1(0))=\gamma_m\left(a_1'\right)\otimes 1_{q_m}. \end{align} $$

Put

(e8.46) $$ \begin{align} c_0'&=\begin{pmatrix} 0 & 0\\ 0 &\gamma_m\left(a_1'\right)\end{pmatrix} \end{align} $$

and

(e8.47) $$ \begin{align} c_0&=\begin{pmatrix} 0 & 0\\ 0 &\gamma_m(a_1(0))\end{pmatrix}\otimes 1_5=\begin{pmatrix} 0 & 0\\ 0 &\gamma_m\left(a_1'\right)\end{pmatrix}\otimes 1_{5q_m}=c_0'\otimes 1_{5q_m}. \end{align} $$

Note that $c_0'\in M_{d_mp_m}$ . Put

(e8.48) $$ \begin{align} c_i(t)&=&\begin{pmatrix} 0 & 0\\ 0 &\gamma_m(a_1\circ \xi_i(t))\end{pmatrix}\otimes 1_5, \quad i=r_0+1,r_0+2,\dotsc,k. \end{align} $$

Recall (see formula (e6.34)) that at $t=0$ ,

(e8.49) $$ \begin{align} \xi_i(0)=\begin{cases} 0 & {\text{if }} 1\le i\le r_0,\\ 1/2 & \text{if } r_0<i\le k. \end{cases} \end{align} $$

Recall also (see the line after equation (e6.35)) that $ r_05q_m=t_0q_{m+1}$ for some integer $t_0\ge 1$ . Hence $\left (c_0'\otimes 1_{5q_m}\right )\otimes 1_{r_0}=c_0'\otimes 1_{r_05q_m}=\left (c_0'\otimes 1_{t_0}\right )\otimes 1_{q_{m+1}}$ . On the other hand, since $k=r_0+m(0)q_{m+1}$ (see equation (e6.30)), we have

(e8.50) $$ \begin{align} \mathrm{diag}\left(c_{r_0+1}(0), \dotsc, c_k(0)\right) =^s\left(\begin{pmatrix} 0 & 0\\ 0 &\gamma_m(a_1(1/2))\end{pmatrix}\otimes 1_5\right)\otimes 1_{m(0)q_{m+1}}. \end{align} $$

Note that $=^s$ is implemented by the same scalar unitary as in equation (e6.36) (see also the end of Notation 6.6 for the notation $=^s$ ). As in formula (e6.37) (and the line after it), since $b_0\in C\!\left([0,1],M_{p_{m+1}q_{m+1}}\right )$ (mentioned earlier), this implies that $b_0\in L_{p_{m+1}, q_{m+1}}$ (see equation (e8.31)).

Since $a_1\ge 0$ , $b_0(t)\le a_1(t)$ for all $t\in [0,1]$ (see equation (e8.44)). Since $\varphi _k$ is an injective unital homomorphism for all k, by equation (e8.40), we also have

(e8.51) $$ \begin{align} \inf\{d_\tau(\varphi_m(a_1)): \tau\in T(A_{m+1})\}=\inf\left\{d_\tau\left(\varphi_{n_1, m+1}(a_0)\right): \tau\in T(A_m)\right\}\ge d. \end{align} $$

By formulas (e8.42), (e8.44), (e8.43) and (e8.41), we conclude that for each $t\in (0,1)$ ,

(e8.52) $$ \begin{align} d_\sigma(e(t))<d_\sigma(b_0(t)),\quad d_{\tau_0}(e(0))<d_{\tau_0}(b_0(0))\quad\text{and}\quad d_{\tau_1}(e(1))<d_{\tau_1}(\varphi_m(a_1)(1)), \end{align} $$

where $\sigma $ is the unique tracial state of $M_{p_{m+1}q_{m+1}}$ , $\tau _0$ is the unique tracial state of $M_{p_{m+1}}\otimes 1_{q_{m+1}}$ and $\tau _1$ is the unique tracial state of $1_{q_{m+1}}\otimes M_{q_{m+1}}$ . Note that $e(1)=0$ . It follows that for all $\tau \in T\left (L_{p_{m+1}, q_{m+1}}\right )$ ,

(e8.53) $$ \begin{align} d_\tau(e)<d_\tau(b_0). \end{align} $$

By, for example, [24, Theorem 3.18],

(e8.54) $$ \begin{align} e\lesssim b_0 \,\,\mathrm{in}\,\, L_{p_{m+1},q_{m+1}}. \end{align} $$

By Lemma 8.2,

(e8.55) $$ \begin{align} e\lesssim \varphi_m(a_1) \mathrm{\ in\ } E_{p_{m+1},q_{m+1}}=A_{m+1}. \end{align} $$

It follows that

(e8.56) $$ \begin{align} e\lesssim \varphi_{m, \infty}(a_1)=f_{1/4}\left(\varphi_{1, \infty}(a')\right)\lesssim a. \end{align} $$

Combining this with (ii) and (iii) here, we conclude that $A_z^C$ is essentially tracially in ${\cal N}_{{\mathscr Z},s,s}$ . Since $B_z\cong {\mathscr Z}$ , which has nuclear dimension $1$ [44, Theorem 6.4], $A_z^C$ has essential tracial nuclear dimension at most $1$ . Since ${\mathscr Z}$ is in ${\cal T}$ , $A_z^C$ is e. tracially in ${\cal T}$ . By Proposition 4.6, every $2$ -quasitrace of $A_z^C$ is a tracial state. By Corollary 5.10, A has stable rank one.

Remark 8.5. Note that the proof of Theorem 8.4 actually shows that $A_z^C$ is essentially tracially approximated by ${\mathscr Z}$ itself, as $B_z\cong {\mathscr Z}$ . Let ${\cal P}$ be the class of separable nuclear $C^*$ -algebras. Then $A_z^C$ is essentially tracially in ${\cal P}$ , since ${\mathscr Z}\in {\cal P}$ . By Proposition 7.4 and Theorem 7.3, $A_z^C$ is not nuclear if C is not nuclear. Since $A_z^C$ has no nonzero projection other than the identity, for nonnuclear C it cannot be TA ${\cal P}$ .

Theorem 8.6. Let $(G, G_+, g)$ be a countable weakly unperforated simple ordered group, F be a countable abelian group, $\Delta $ be a metrisable Choquet simplex and $\lambda : G\to \operatorname {Aff}_+(\Delta )$ be a homomorphism.

Then there is a unital simple nonexact (or exact but nonnuclear) $C^*$ -algebra A which is e. tracially in ${\cal N}_{{\mathscr Z}, s,s}$ and has essential tracial nuclear dimension at most $1$ , stable rank one and strict comparison for positive elements, such that

(e8.57) $$ \begin{align} (K_0(A), K_0(A)_+, [1_A], K_1(A), T(A), \rho_A)=(G, G_+, g, F, \Delta, \lambda). \end{align} $$

Proof. It follows from [24, Theorem 13.50] that there is a unital simple $A_0$ with finite nuclear dimension which satisfies the UCT such that

(e8.58) $$ \begin{align} \left(K_0(A_0), K_0(A_0)_+, \left[1_{A_0}\right], K_1(A_0), T(A_0), \rho_{A_0}\right)=(G, G_+, g, F, \Delta, \lambda). \end{align} $$

Let $A_z^C$ be the $C^*$ -algebra in Theorem 7.3 with $A_z^C$ nonexact (or exact but nonnuclear). Then define $A:= A_0\otimes A_z^C$ . Note that since $A_0$ is a separable amenable $C^*$ -algebrasatisfying the UCT, by [45, Theorem 2.14], the Künneth formula holds for the tensor product $A=A_0\otimes A_z^C$ . Since the only normalised $2$ -quasitrace of $A_z^C$ is the unique tracial state, and

$$ \begin{align*} \left(K_0\left(A_z^C\right), K_0\left(A_z^C\right)_+, \left[1_{A_z^C}\right], K_1\left(A_z^C\right)\right)=(\mathbb{Z}, \mathbb{Z}_+, 1, 0), \end{align*} $$

one computes (applying the Künneth formula) that

(e8.59) $$ \begin{align} (K_0(A), K_0(A)_+, [1_{A}], K_1(A), T(A), \rho_{A})=(G, G_+, g, F, \Delta, \lambda). \end{align} $$

We will show that A is essentially tracially in ${\cal N}_{{\mathscr Z},s,s}$ and has e. tracial nuclear dimension at most $1$ . Once this is done, by Definition 8.3 A has essentially tracial nuclear dimension at most $1$ and by Corollary 5.10 has stable rank one and strict comparison for positive elements.

To see that A is essentially tracially in ${\cal N}_{{\mathscr Z},s,s}$ , set ${{1>\varepsilon }}>0$ , let ${\cal F}\subset A^{\textbf{1}}\setminus \{0\}$ be a finite subset and set $d\in A_+\setminus \{0\}$ . Without loss of generality, we may assume that $\lVert x\rVert \ge \varepsilon $ for all $x\in {\cal F}$ . For each $x\in {\cal F}$ , put $a(x)=g_ {{\varepsilon ^2/32}} (xx^*)$ and $b(x)=g_{{\varepsilon ^2/16}} (xx^*)$ , where $g_r(t)\in C_0([0,1])$ such that $g_r(t)=0$ for $t\in \left [0, \lVert x\rVert ^{{2}}-r/2\right ]$ , $g_r(t)=1$ for $t\in \left [\lVert x\rVert ^{{2}}-r/4, 1\right ]$ and $g_r$ is linear in $\left [\lVert x\rVert ^{{2}}-r/2, \lVert x\rVert ^{{2}}-r/4\right ]$ . Note that $b(x)xx^*b(x)=\left (g_{{\varepsilon ^2/16}} ( xx^*)\right )^2xx^*$ . Therefore,

(e8.60) $$ \begin{align} b(x)xx^*b(x)\ge \left(\lVert xx^*\rVert-(\varepsilon/4)^2\right)b(x)^2. \end{align} $$

By Kirchberg’s slice lemma (see, for example, [41, Lemma 4.1.9]), for each $x\in {\cal F}$ there are $c(x)\in (A_0)_+\setminus \{0\}$ , $d(x)\in {A_z^C}_+\setminus \{0\}$ and $z(x)\in A_0\otimes A_z^C$ such that $z(x)^*z(x)=c(x)\otimes d(x)$ and $z(x)z(x)^*\in \mathrm {Her}(a(x))$ . We may assume that $\lVert c(x)\rVert =\lVert d(x)\rVert =1$ . This also implies that $\lVert z(x)z(x)^*\rVert =1$ . Put

$$ \begin{align*} {\cal F}':={\cal F}\cup\left\{x^*x, xx^*, a(x), b(x), c(x), d(x), d(x)^{1/2}, z(x), z^*(x), z(x)^*z(x), z(x)z(x)^*: x\in {\cal F}\right\}. \end{align*} $$

Without loss of generality, we may assume that there are $n\in \mathbb {N}$ , $M\ge 1$ and finite subsets ${\cal F}_0\subset A$ and ${\cal F}_1\subset A_z^C$ such that for all $y\in {\cal F}'$ ,

(e8.61) $$ \begin{align} y\in_{\varepsilon^2/64} {\cal F}'':= \left\{\sum_{i=1}^na_i\otimes b_i, a_i\in {\cal F}_0, b_i\in {\cal F}_1\right\}, \end{align} $$

(e8.62) $$ \begin{align} c(x)\in {\cal F}_0,\ d(x), d(x)^{1/2}\in {\cal F}_1\mathrm{\ for\ all\ } x\in {\cal F}, \end{align} $$

(e8.63) $$ \begin{align} \lVert f_0\rVert, \ \lVert f_1\rVert\le M\text{if } f_0\in {\cal F}_0\mathrm{\ and\ } f_1\in {\cal F}_1. \end{align} $$

By Kirchberg’s slice lemma, there are $a_0\in (A_0)_+\setminus \{0\}$ and $b_0\in \left (A_z^C\right )_+\setminus \{0\}$ such that $a_0\otimes b_0\lesssim d$ .

Let us identify $A_0$ with $A_0\otimes {\mathscr Z}$ (see [48, Corollary 7.3]). In $A_0\otimes {\mathscr Z}$ , choose $1_{A_0}\otimes a_z$ with $a_z\in {\mathscr Z}_+\setminus \{0\}$ such that $1_{A_0}\otimes a_z\lesssim _{A_0} a_0$ . Choose $b_z\in (B_z)_+\setminus \{0\}\subset A_z^C$ such that $b_z\lesssim _{A_z^C} b_0$ .

Note that $B_z\cong {\mathscr Z}_b\otimes B_z$ , where ${\mathscr Z}_b\cong {\mathscr Z}$ . Put $c_0:=\sigma (a_z)\otimes b_z\in B_z$ , where $\sigma : 1_{A_0}\otimes {\mathscr Z}(\subset A_0)\to {\mathscr Z}_b\otimes 1_{B_z}$ is an isomorphism. Consider $D_0:=A_0\otimes \sigma \left (1_{A_0}\otimes {\mathscr Z}\right )\otimes 1_{B_z}\subset A_0\otimes B_z$ . We may also write $D_0=(A_0\otimes {\mathscr Z})\otimes \sigma \left (1_{A_0}\otimes {\mathscr Z}\right )\otimes 1_{B_z}$ . There is a sequence of unitaries $v_n\in D_0$ such that

(e8.64) $$ \begin{align} \lim_{n\to\infty}v_n^*\left(1_{A_0}\otimes \sigma(a_z)\otimes 1_{B_z}\right)v_n=1_{A_0}\otimes a_z\otimes 1_{B_z}. \end{align} $$

It follows that

(e8.65) $$ \begin{align} 1_{A_0}\otimes c_0=1_{A_0}\otimes \sigma(a_z)\otimes b_z \sim {{1_{A_0}\otimes a_z\otimes b_z}}\lesssim a_0\otimes b_0\lesssim d. \end{align} $$

By Remark 8.5, there exists $e_1\in A_z^C$ with $0\le e_1\le 1$ such that for all $f\in {\cal F}_1$ ,

(e8.66) $$ \begin{align} \lVert e_1f-fe_1\rVert&<\varepsilon^2/ {{64}} {{(nM)}}, \qquad\left(1_{A_z^C}-e_1\right)f\in_{\varepsilon^2/ {{64}} {{(nM)}}} B_z, \end{align} $$

(e8.67) $$ \begin{align} \left\lVert\left(1_{A_z^C}-e_1\right)f\right\rVert&\ge \left(1-\varepsilon^2/{{ {{64}} (nM)}}\right)\lVert f\rVert \mathrm{\ and\ } e_1\lesssim c_0. \end{align} $$

Put $B=A_0\otimes B_z$ . Then by [10, Theorem B], B is a separable simple ${\mathscr Z}$ -stable $C^*$ -algebra and has nuclear dimension at most $1$ .

Put $e=1_{A_0}\otimes e_1$ . Then $0\le e\le 1$ . For any $f''\in {\cal F}''$ , we have that $f''=\sum _{i=1}^n a_i\otimes b_i$ for some $a_i\in {\cal F}_0$ and $b_i\in {\cal F}_1$ . It follows from formula (e8.66) that

(e8.68) $$ \begin{align} \lVert ef''-f''e\rVert=\left\lVert e\left(\sum_{i=1}^n a_i\otimes b_i\right)-\left(\sum_{i=1}^n a_i\otimes b_i\right)e\right\rVert =\left\lVert\sum_{i=1}^n a_i\otimes (e_1b_i-b_ie_1)\right\rVert<\varepsilon^2/ {{64}}. \end{align} $$

Also by formula (e8.66), for $f''\in {\cal F}''$ ,

(e8.69) $$ \begin{align} &&\hspace{-0.2in}(1-e)f''=\left(1-1_{A_0}\otimes e_1\right)\left(\sum_{i=1}^n a_i\otimes b_i\right)=\sum_{i=1}^na_i\otimes \left(1_{A_z^C}-e_1\right)b_i\in_{\varepsilon^2/ {{64}} } A_0\otimes B_z. \end{align} $$

It follows (recall formula (e8.61)) that for all $f\in {\cal F}$ ,

(e8.70) $$ \begin{align} \lVert ef-fe\rVert<{{\varepsilon^2}}/16\qquad\text{and}\qquad (1-e)f\in_{{{\varepsilon^2/16}}}A_0\otimes B_z=B. \end{align} $$

Moreover, by formulas (e8.65) and (e8.67),

(e8.71) $$ \begin{align} e\lesssim a_0\otimes b_0\lesssim d. \end{align} $$

To estimate $\lVert (1-e)x\rVert $ for $x\in {\cal F}$ , we note that by formula (e8.67) (recall that $\lVert c(x)\rVert =\lVert d(x)\rVert =1$ ), for $x\in {\cal F}$ we have

(e8.72) $$ \begin{align} \left\lVert(1-e)(c(x)\otimes d(x))^{1/2}\right\rVert &= \left\lVert c(x)^{1/2}\otimes \left(1_{A_z^C}-e_1\right)d(x)^{1/2}\right\rVert \hspace{4pc} \end{align} $$

(e8.73) $$ \begin{align} & \hspace{11pc} \ge \left(1-\varepsilon^2/{{64}}\right)\lVert c(x)\rVert\lVert d(x)\rVert=\left(1-\varepsilon^2/{{64}}\right). \end{align} $$

Then by formulas (e8.70) and (e8.73),

(e8.74) $$ \begin{align} \nonumber \left\lVert(1-e)z(x)z(x)^*(1-e)\right\rVert&=\left\lVert z(x)^*(1-e)^2z(x)\right\rVert\\[2pt] &\ge \left\lVert(1-e)z(x)^*z(x)(1-e)\right\rVert-\varepsilon^2/{{8}}\\[2pt]\nonumber &=\left\lVert(1-e)(c(x)\otimes d(x))^{1/2}\right\rVert^2-\varepsilon^2/{{8}} \end{align} $$

(e8.75) $$ \begin{align} & \hspace{9.5pc} \ge\left(1-\varepsilon^2/{{64}}\right)^2-\varepsilon^2/{{8}}>{{1-5\varepsilon^2/32}}. \end{align} $$

Since $b(x)z(x)z(x)^*=z(x)z(x)^*b(x)=z(x)z(x)^*$ , we compute by formulas (e8.60) and (e8.75)) that for all $x\in {\cal F}$ ,

(e8.76) $$ \begin{align} \left\lVert(1-e)xx^*(1-e)\right\rVert&\ge \left\lVert(1-e)b(x)^{1/2}xx^*b(x)^{1/2}(1-e)\right\rVert \hspace{7pc} \end{align} $$

(e8.77) $$ \begin{align} &\ge {{\left(\lVert x\rVert^2-\varepsilon^2/16\right)}} \lVert(1-e)b(x)(1-e)\rVert \hspace{1.8pc} \end{align} $$

(e8.78) $$ \begin{align} &\ge {{\left(\lVert x\rVert^2-\varepsilon^2/16\right)}}\left\lVert(1-e)z(x)z(x)^*(1-e)\right\rVert \end{align} $$

(e8.79) $$ \begin{align} & \hspace{1.7pc} \ge \left(\lVert x\rVert^2-\varepsilon^2/16\right)\left(1-{{5\varepsilon^2/32}}\right)\ge \lVert x\rVert^2-{{7\varepsilon^2/32}} \end{align} $$

(e8.80) $$ \begin{align} & \ge \lVert x\rVert^2 -2\varepsilon\lVert x\rVert+\varepsilon^2. \hspace{7.7pc} \end{align} $$

(Recall for the last inequality that $\lVert x\rVert \ge \varepsilon $ .) It follows that for all $x\in {\cal F}$ ,

(e8.81) $$ \begin{align} \lVert(1-e)x\rVert\ge \lVert x\rVert-\varepsilon. \end{align} $$

This, together with formula (e8.70), implies that $A_0\otimes A_z^C$ is essentially tracially in ${\cal N}_{{\mathscr Z},s,s}$ , since B is ${\mathscr Z}$ -stable and has nuclear dimension no more than $1$ (see [10, Theorem B]).

Now suppose that we choose $A_z^C$ not exact. Since $A_z^C$ embeds into $A_0\otimes A_z^C$ and $A_z^C$ is not exact, $A_0\otimes A_z^C$ is also not exact (see, for example, [47, Proposition 2.6]). If $A_z^C$ is exact but not nuclear, then $A_0\otimes A_z^C$ is exact but not nuclear (by [9, Propositions 10.2.7, 10.1.7]).

Remark 8.7.

1. (1) Let $A_0$ be a unital separable nuclear purely infinite simple $C^*$ -algebra in the UCT class. Then the proof of Theorem 8.6 also shows that $A:=A_0\otimes A_z^C$ is a nonexact purely infinite simple $C^*$ -algebra (if C is nonexact) which has essential tracial nuclear dimension $1$ and $\mathrm {Ell}(A)=\mathrm {Ell}(A_0)$ .

2. (2) If the RFD $C^*$ -algebra C at the beginning of Section 6 is amenable, then $C_0((0,1], C)$ is a nuclear contractible $C^*$ -algebra which satisfies the UCT. It follows that the unitisation B of $C_0((0,1], C)$ also satisfies the UCT. Therefore $D(m,k)$ and $I=C_0((0,1), M_{mk}(B))$ in formula (e6.8) satisfy the UCT. Thus $E_{m,k}$ is nuclear and satisfies the UCT. It follows that $A_z^C$ is a unital amenable separable simple stable rank one $C^*$ -algebra with a unique tracial state which also has strict comparison for positive elements and satisfies the UCT. By [34, Theorem 1.1], $A_z^C$ is ${\mathscr Z}$ -stable. By [35, Theorem 1.1], $A_z^C$ has finite nuclear dimension. Then by [16, Corollary 4.11], $A_z^C$ is classifiable by the Elliott invariant (see also [16, Remark 4.6]). Since $A_z^C$ has the same Elliott invariant of ${\mathscr Z}$ , it follows that $A_z^C\cong {\mathscr Z}$ .

3. (3) On the other hand, we make no attempt at this time to classify $C^*$ -algebras $A_z^C$ constructed in Section 6 in the nonnuclear cases. We do not know whether $A_z^{C_1}$ is isomorphic to $A_z^{C_2}$ if $C_1$ and $C_2$ are nonisomorphic, nonnuclear $C^*$ -subalgebras. In fact, as it stands, $A_z^C$ may depend on the connecting maps used in the construction.