Proof. Suppose that η is continuous. It remains to show that η ⁻¹ is continuous. We show that η maps closed sets to closed sets. By [Reference Palais43, corollary], it suffices to check that η is a proper map i.e. the preimage of any compact set is compact. Let $K\subseteq S$ be compact and note that $\pi(K)$ is compact since π is continuous. Since $\pi\circ\eta=\pi'$, it follows that

\begin{align*}\eta^{-1}(K)\subseteq \eta^{-1}(\pi^{-1}(\pi(K)))=(\pi')^{-1}(\pi(K)).\end{align*}

But $(\pi')^{-1}(\pi(K))$ is compact since $(S',\pi')$ is a proper simplex bundle (see definition 1.3), so $\eta^{-1}(K)$ is compact as a closed subset of a compact set. This finishes the argument. The case when η ⁻¹ is assumed to be continuous is completely symmetric.

Proof. We first note that $B\otimes\mathcal{K}\cong A\otimes\mathcal{K}$ is nuclear and $\mathcal{Z}$-stable. Since nuclearity passes to hereditary subalgebras, B is nuclear. Moreover, B is $\mathcal{Z}$-stable by [Reference Toms and Winter58, corollary 3.1]. Finally, the UCT is closed under stable isomorphisms ([Reference Blackadar4, 22.3.5 (a)]), so B satisfies the UCT.

Proof. Note that $e\left(B\rtimes_\gamma\mathbb{Z}\right)e$ is unital as e is a projection, and it is separable since B is separable and $\mathbb{Z}$ is a countable discrete group. If $B\rtimes_\gamma\mathbb{Z}$ is simple and e is full, then $e\left(B\rtimes_\gamma\mathbb{Z}\right)e$ is simple. If B is nuclear, then $B\rtimes_\gamma\mathbb{Z}$ is nuclear by [Reference Brown and Ozawa11, theorem 4.2.4]. Since nuclearity passes to hereditary subalgebras, $e\left(B\rtimes_\gamma\mathbb{Z}\right)e$ is nuclear. Suppose now that B satisfies the UCT. Since the $\mathrm{UCT}$ is closed under crossed products by $\mathbb{Z}$ ([Reference Blackadar4, 22.3.5 (g)]), $B\rtimes_\gamma\mathbb{Z}$ satisfies the $\mathrm{UCT}$. Furthermore, $B\rtimes_\gamma\mathbb{Z}$ and $e\left(B\rtimes_\gamma\mathbb{Z}\right)e$ are stably isomorphic (see [Reference Brown10]) and the $\mathrm{UCT}$ is closed under stable isomorphisms ([Reference Blackadar4, 22.3.5 (a)]). Hence, $e\left(B\rtimes_\gamma\mathbb{Z}\right)e$ satisfies the $\mathrm{UCT}$. Since $\mathcal{Z}$-stability passes to hereditary subalgebras ([Reference Toms and Winter58, corollary 3.1]) and $B\rtimes_\gamma\mathbb{Z}$ is $\mathcal{Z}$-stable, it follows that $e\left(B\rtimes_\gamma\mathbb{Z}\right)e$ is $\mathcal{Z}$-stable.

Proof. Let $(G,G_+,v, H, X, \rho)$ be as in the statement of the theorem. Then there exists a simple separable stable nuclear $\mathrm{C}^*$-algebra B such that $\mathrm{Ell}(B)\cong (G,G_+,H, \tilde{X},\tilde{\rho})$, where $\tilde{X}$ is the cone with base X and $\tilde{\rho}$ is the unique extension of the pairing ρ ([Reference Elliott22, theorem 5.2.3.2]). Since the $\mathrm{C}^*$-algebra B is constructed as an inductive limit, one can simply tensor with $\mathcal{Z}$ at each stage to assume that B is $\mathcal{Z}$-stable, or notice that B is simple and has finite nuclear dimension as each building block has finite nuclear dimension ([Reference Winter and Zacharias61, proposition 2.3]). Thus, B is $\mathcal{Z}$-stable by [Reference Tikuisis56, corollary 8.7]. Moreover, the UCT is preserved by inductive limits (see [Reference Blackadar4, 22.3.5 (e)]), so B satisfies the UCT.

If $p\in B$ is a projection such that $[p]_0=v$ in $K_0(B)$, then we take A = pBp. Since B is simple, p is a full projection, so A is simple separable and unital (see [Reference Brown10]). Moreover, A is nuclear, $\mathcal{Z}$-stable, and it satisfies the UCT by proposition 2.10. Finally, it is immediate to see that $\mathrm{Ell}(A)\cong (G,G_+,v, H, X, \rho).$

Proof. This follows from [Reference Goodearl31, theorem 12.3] (or [Reference Blackadar and Handelman5, theorem III.1.3]) since any lower semicontinuous extended quasitrace on an A$\mathbb{T}$-algebra is an extended trace ([Reference Brown and Winter12, theorem 6]).

Proof. Using the strategy in [Reference Thomsen54, lemma 3.4], we will adapt Liao’s proof of [Reference Liao41, proposition 2.3]. Let $P:A\rtimes_\gamma\mathbb{Z}\to A$ be the canonical conditional expectation. Given a densely defined lower semicontinuous trace τ on $A\rtimes_\gamma\mathbb{Z}$, we will show that $\tau=\tau|_A\circ P$. Since q_r is a projection and hence contained in the Pedersen ideal of A, we have that $\tau(q_r) \lt \infty$. Moreover,

\begin{align*}\tau(x)=\lim\limits_{r\to\infty}\tau(q_rxq_r)\end{align*}

and

\begin{align*}\tau(P(x))=\lim\limits_{r\to\infty}\tau(q_rP(x)q_r)\end{align*}

for any positive element x in $A\rtimes_\gamma\mathbb{Z}$. Therefore, to show that $\tau=\tau|_A\circ P$, it suffices to show that

\begin{align*}\tau(q_rxq_r)=\tau(q_rP(x)q_r)\end{align*}

for any positive element x in $A\rtimes_\gamma\mathbb{Z}$ and any $r\in\mathbb N$. Since τ is nonzero, we can assume that $\tau(q_r) \gt 0$ for any $r\in\mathbb{N}$.

Let u be the unitary in $\mathcal{M}(A\rtimes_\gamma\mathbb{Z})$ implementing the action. Since the map $x\mapsto \tau(q_rxq_r)$ is bounded on $A\rtimes_\gamma\mathbb{Z}$ with norm $\tau(q_r)$, it suffices to show that $\tau(q_rau^nq_r)=0$ for any $a\in A$, and any $r,n\in\mathbb{N}$.

Recall that the automorphism γ has finite Rokhlin dimension, say $d\in\mathbb{N}$. Let $r,n\in \mathbb{N}$, $a\in A$ be a contraction, and ϵ > 0. Let $p = 2n$ and $\epsilon' = \frac{\epsilon}{(6(d+1)n+1)\tau(q_r)}$. Using (ii) of definition 1.16, it follows that there exist $\{f_{k}^{(\ell)}:0\leq k\leq p-1, \;0\leq \ell \leq d \}$ in A such that

(1.1)\begin{align} \left\| \left(\sum_{\ell=0}^d \sum_{k=0}^{p-1} f_k^{(\ell)}\right)a - a \right\| \lt \epsilon'. \end{align}

Combining (i), (iii), and (iv) of definition 1.16, we can further assume that

(1.2)\begin{align} \| {f_k^{(\ell)} }^{\frac{1}{2} } a \gamma^n( {f_k^{(\ell)} }^{\frac{1}{2}} ) \| \lt \epsilon'\;\;\;\; (\ell=0,1,...,d;\; k=0,...,p-1). \end{align}

Moreover, by (iii) of definition 1.16, we can ensure that

(1.3)\begin{align} \|{f_k^{(l)}}^{\frac{1}{2}}q_r-q_r{f_k^{(l)}}^{\frac{1}{2}}\| \lt \epsilon' \end{align}

for any $0\leq k\leq p-1$ and any $0\leq l\leq d$.

Then

(1.4)\begin{align} | \tau(q_rau^nq_r) | \stackrel{(1.1)}{\leq} \left| \sum_{\ell=0}^d\sum_{k=0}^{p-1} \tau(q_rf_k^{(\ell)}au^nq_r) \right| + \tau(q_r)\epsilon'. \end{align}

Fix $k,l$ and let $x:= {f_k^{(l)}}^{\frac{1}{2}}au^n$. Using that τ is a trace and that x is a contraction, we get that

(1.5)\begin{align} \begin{array}{rcl} |\tau(q_rx(q_r{f_k^{(l)}}^{\frac{1}{2}}-{f_k^{(l)}}^{\frac{1}{2}}q_r))|&=& |\tau(q_rx(q_r{f_k^{(l)}}^{\frac{1}{2}}-{f_k^{(l)}}^{\frac{1}{2}}q_r)q_r)|\\ &\leq&\tau(q_r)\|x(q_r{f_k^{(l)}}^{\frac{1}{2}}-{f_k^{(l)}}^{\frac{1}{2}}q_r)\|\\ &\stackrel{(1.3)}{\leq}& \tau(q_r)\epsilon'. \end{array} \end{align}

Similarly, one has that

\begin{align*} |\tau({f_k^{(\ell)}}^{\frac{1}{2}}q_r{f_k^{(\ell)}}^{\frac{1}{2}}au^nq_r)-\tau(q_rf_k^{(\ell)}au^nq_r)| &=|\tau(q_r({f_k^{(\ell)}}^{\frac{1}{2}}q_r-q_r{f_k^{(\ell)}}^{\frac{1}{2}})xq_r)|\\ &\leq \tau(q_r)\epsilon'. \end{align*}

Therefore, using the last two estimations above, we get that

\begin{align*} &|\tau(q_r{f_k^{(\ell)}}^{\frac{1}{2}} a u^n {f_k^{(\ell)}}^{\frac{1}{2}}q_r)-\tau(q_rf_k^{(\ell)}au^nq_r)|\\ &\leq |\tau(q_r{f_k^{(\ell)}}^{\frac{1}{2}}au^nq_r{f_k^{(\ell)}}^{\frac{1}{2}})-\tau(q_rf_k^{(\ell)}au^nq_r)|+\tau(q_r)\epsilon'\\ &= |\tau({f_k^{(\ell)}}^{\frac{1}{2}}q_r{f_k^{(\ell)}}^{\frac{1}{2}}au^nq_r)-\tau(q_rf_k^{(\ell)}au^nq_r)|+\tau(q_r)\epsilon'\\ &\leq 2\tau(q_r)\epsilon'. \end{align*}

Thus, from (1.4) we now get that

\begin{align*}|\tau(q_rau^nq_r)|\leq \sum_{\ell=0}^d\sum_{k=0}^{p-1} \left| \tau(q_r{f_k^{(\ell)}}^{\frac{1}{2}} a u^n {f_k^{(\ell)}}^{\frac{1}{2}}q_r) \right| +(2(d+1)p+1)\tau(q_r)\epsilon'\end{align*}

\begin{align*} \begin{array}{rcl} &=&\sum_{\ell=0}^d\sum_{k=0}^{p-1} \left| \tau(q_r{f_k^{(\ell)}}^{\frac{1}{2}} a \gamma^n( {f_k^{(\ell)}}^{\frac{1}{2}} ) u^n q_r) \right| + (2(d+1)p+1)\tau(q_r)\epsilon' \\ &\stackrel{(1.2)}{\leq}&((d+1)p + 2(d+1)p+1)\tau(q_r)\epsilon' \\ &=&(6(d+1)n+1)\tau(q_r)\epsilon' = \epsilon. \end{array} \end{align*}

Hence, $\tau(q_rau^nq_r)=0$ for any $r,n\in\mathbb{N}$, which yields that $\tau(q_rxq_r)=\tau(q_rP(x)q_r)$ for any $r\in\mathbb{N}$ and any $x\in A\rtimes_\gamma\mathbb{Z}$. Therefore, the canonical restriction induces an injection from the space of densely defined lower semicontinuous traces on $A\rtimes_{\gamma}\mathbb{Z}$ into the space of γ-invariant densely defined lower semicontinuous traces on A. Since this restriction is also surjective, the conclusion follows.

Proof. The construction of the pair $(G,G_+)$ is the one provided in [Reference Elliott, Sato and Thomsen26, theorem 5.1]. Since the topology on S is second countable, we can choose a countable subgroup G ₀ of $\mathcal{A}(S,\pi)$ satisfying (i) in the statement of the lemma. Moreover, we can ensure that

• • The function

  \begin{align*}x\mapsto e^{n\pi(x)}(1-e^{-\pi(x)})^mf(x)\end{align*}

  is in G ₀ for any $f\in G_0$ and any $n,m\in\mathbb{Z}$;

• • The functions

  \begin{align*}x\mapsto (\chi_{(-\infty,0]}\circ\pi)(x)e^{n\pi(x)}(1-e^{-\pi(x)})^m\end{align*}

  and

  \begin{align*}x\mapsto (\chi_{[0,\infty)}\circ\pi)(x)e^{n\pi(x)}(1-e^{-\pi(x)})^m\end{align*}

  are in G ₀ for any $n,m\in\mathbb{Z}$. The functions $\chi_{(-\infty,0]}\circ\pi$ and $\chi_{[0,\infty)}\circ\pi$ are continuous on S because $\pi^{-1}(0)=\emptyset$.

Let $G=\mathbb{Q}G_0$ and $G_+=\{f\in G: f(x) \gt 0, x\in S\}\cup\{0\}.$

That $(G,G_+)$ is a dimension group satisfying the strong Riesz interpolation property follows from [Reference Elliott, Sato and Thomsen26, lemma 5.3]. Since the function

\begin{align*}x\mapsto e^{n\pi(x)}(1-e^{-\pi(x)})^mf(x)\end{align*}

is in G ₀ for any $f\in G_0$ and any $n,m\in\mathbb{Z}$, we get that α is an automorphism of G and $\alpha(G_+)=G_+$. Furthermore, note that $\pi^{-1}(0)=\emptyset$, so $\mathrm{id}-\alpha$ is also an automorphism of G. This proves (iii). Condition (iv) follows from the proof of [Reference Elliott, Sato and Thomsen26, lemma 5.5], while the existence of an element $\omega\in\pi^{-1}(\beta)$ as in (v) is shown in [Reference Elliott, Sato and Thomsen26, lemma 5.6] (note that both the group G and the map ψ appearing in the proof of [Reference Elliott, Sato and Thomsen26, lemma 5.6] coincide with the ‘G’ and ‘ψ’ in the statement of the lemma).

Suppose that there exists $\omega'\neq \omega$ in $\pi^{-1}(\beta)$ such that $g(\omega)=g(\omega')$ for all $g\in G$. Then, by (i) above, $h(\omega)=h(\omega')$ for any $h\in \mathcal{A}(S,\pi)$ with compact support. As $\omega,\omega'\in \pi^{-1}(\beta)$, there exists $f\in \mathrm{Aff}(\pi^{-1}(\beta))$ such that $f(\omega)\neq f(\omega')$. Therefore, there exist $n\in\mathbb{N}$ and $h\in\mathcal{A}(S,\pi)$ such that the support of h is contained in $\pi^{-1}([-n,n])$ and $h(\omega)\neq h(\omega')$, which is a contradiction. Hence, ω is unique.

Proof. We apply [Reference Rørdam47, proposition 3.5] to the pairs $(K_0(A),0)$ and $(K_1(A),0)$. Moreover, if $H_0=\{0\}$, which happens only when $K_0(A)=0$, we take $H_0=\mathbb{Q}$ and $\kappa_0(x)=2x$.

Proof. We first check (i). As $(G,G_+)$ is a dimension group satisfying the strong Riesz interpolation property (lemma 3.1(ii)) and H ₀ is nonzero and torsion free, $(\Gamma, \Gamma_+)$ is a dimension group by [Reference Effros, Handelman and Shen19, lemma 3.2].

To check (ii), it suffices to find $h\in \Gamma_+$ such that $2h\leq g\leq 3h$. We then let $h_1=3h-g$ and $h_2=g-2h$. Let $g\in \Gamma_+$ be nonzero. Since $G=\mathbb{Q}G$, there exists $h'\in G_+$ nonzero such that $\frac{7}{3}h'\leq p(g)\leq \frac{8}{3}h'$. Then, $p(g-2(0,h'))=p(g)-2h' \gt 0$ and $p(3(0,h')-g)=3h'-p(g) \gt 0$, so we can take $h=(0,h')$ to conclude that $2h\leq g\leq 3h$ in Γ.

We now check (iii). Let I be a nonzero order ideal of Γ. Since $I=\left(I\cap \Gamma_+\right)-\left(I\cap\Gamma_+\right)$, there exist $h'\in H_0$ and $g\in G_+\setminus\{0\}$ such that $(h',g)\in I$. Then, for any $h\in H_0$, one has that

\begin{align*}0 \lt (h+h',g) \lt 2(h',g)\end{align*}

in Γ. Since I is an order ideal, it follows that $(h+h',g)\in I$. Combining this with the fact that $(h',g)\in I$, yields that $(h,0)\in I$. Hence $H_0\oplus 0\subseteq I$. If $g\in p(I)$, there exists $h'\in H_0$ such that $(h',g)\in I$. If $h\in H_0$, then $(h+h',0)\in I$, so $(h,g)\in I$. Hence, we get that $I=H_0\oplus p(I)$.

The proof of (iv) is contained in [Reference Elliott, Sato and Thomsen26, lemma 5.5], but we will include the details for the convenience of the reader. Let I be a nonzero order ideal of Γ such that $(\kappa_0\oplus\alpha)(I)=I$. By condition (iii), it follows that $I=H_0\oplus p(I)$. Moreover, p(I) is an order ideal in G such that $\alpha(p(I))=p(I)$. Condition (iv) of lemma 3.1 yields that $G=p(I)$. Thus, $I=\Gamma$.

We prove (v) by contradiction. Let $J\subseteq I$ be order ideals of Γ such that $I/J$ is nonzero and liminary in the category of ordered groups i.e. any prime quotient of $I/J$ is isomorphic to $\mathbb{Z}$. Then, by (iii) above, $I=H_0\oplus p(I)$. Suppose that there exists a surjection $q:(H_0\oplus p(I))/J\to\mathbb{Z}$ and let (h, i) be positive in $H_0\oplus p(I)$ such that $q(h,i)=1$. As $p(I)\subseteq G$ is an order ideal, it follows that $\frac{1}{2}i\in p(I)$. If $(h,\frac{1}{2}i)\in J$, then $(0,\frac{1}{2}i)\in J$ by (iii). Therefore, $(h,i)\in J$, which is a contradiction, so $(h,\frac{1}{2}i)\notin J$. Hence, one gets that

\begin{align*}0\leq q(h,\frac{1}{2}i)\leq 1.\end{align*}

If $q(h,\frac{1}{2}i)=0$, then $q(4h,2i)=0$. But $(h,i)\leq (4h,2i)$, so $1=q(h,i)\leq 0$, which is a contradiction. If $q(h,\frac{1}{2}i)=1$, then $q(0,\frac{1}{2}i)=0$, which gives that $q(0,2i)=0$. But $(h,i)\leq (0,2i)$, so $1=q(h,i)\leq 0$, which is a contradiction. Thus, the subquotient $I/J$ is not liminary.

Proof. This follows as in [Reference Elliott, Sato and Thomsen26, lemma 5.6]. Let $h\in H_0$ and $n\geq 1$. Then $n(h,0)+v\in \Gamma_+$ and $-n(h,0)+v\in \Gamma_+$, which yields that $n\varphi(h,0)+1\geq 0$ and $-n\varphi(h,0)+1\geq 0$ for all $n\geq 1$. Hence, $\varphi(h,0)=0$, which yields a positive homomorphism $\psi:G\to\mathbb{R}$ such that $\psi\circ p=\varphi$. Since $\varphi(v)=1$ and $p(v)=1$, it follows that $\psi(1)=1$. Moreover, $\varphi\circ(\kappa_0\oplus\alpha)=e^{-\beta}\varphi$ and $p\circ(\kappa_0\oplus\alpha)(g)=\alpha(g)$ for any $g\in G$, so $\psi\circ\alpha=e^{-\beta}\psi$. Thus, (v) of lemma 3.1 shows that there is a unique $\omega\in \pi^{-1}(\beta)$ such that $\psi(g)=g(\omega)$ for all $g\in G$. Hence, $\varphi(h,g)=g(\omega)$ for all $(h,g)\in \Gamma$. The last statement follows by taking β = 0, as $\pi^{-1}(0)=\emptyset$.

Proof. Conditions (i) and (iii) follow as a direct application of [Reference Elliott and Gong23, theorem 4.28]. It is worth noting that in [Reference Elliott and Gong23], the term ideal stands for what is called order ideal in this paper (see [Reference Elliott and Gong23, remark 4.27]).

We first need to define a map from order ideals of Γ to subgroups of $H_1\oplus G$. If I is a nonzero order ideal of Γ, then $I=H_0\oplus p(I)$ by (iii) of lemma 3.3. We then consider the map from order ideals of Γ to subgroups of $H_1\oplus G$ given by $H_0\oplus p(I)\mapsto H_1\oplus p(I)$ which maps 0 to 0. This map also preserves inclusions, upward directed unions, and maps Γ into $H_1\oplus G$. Condition (i) now follows from [Reference Elliott and Gong23, theorem 4.28].

We now prove (ii). Let $((h_0,g_0),(h_1,g_1))\in \Lambda_*$ and $n\in\mathbb{N}$ such that $n((h_0,g_0),(h_1,g_1))\in (\Lambda_*)_+$. Recall from proposition 3.2 that H ₀ and H ₁ are torsion free. Then $\Lambda_*$ is torsion free, so we can assume that $(h_0,g_0)\in \Gamma\setminus\{0\}$. By (i) of lemma 3.3, Γ is a dimension group and hence unperforated, so $(h_0,g_0)\in\Gamma_+$. Moreover, since $n((h_0,g_0),(h_1,g_1))\in (\Lambda_*)_+$, we have that ng ₁ is in the order ideal of G generated by ng ₀, say I ₀. As g ₀ is positive, it follows that $g_0\in I_0$, so I ₀ is generated by g ₀. Since $ng_1\in I_0$, there exists $k\in\mathbb{N}$ such that

\begin{align*}-kng_0\leq ng_1\leq kng_0\end{align*}

in I ₀. Recall from lemma 3.1 that $G=\mathbb{Q}G$. In particular, G is divisible, so it follows that $-kg_0\leq g_1\leq kg_0$, which implies that $0\leq g_1+kg_0\leq 2kg_0$. This yields that $g_1+kg_0\in I_0$ and so $g_1\in I_0$. Thus $((h_0,g_0),(h_1,g_1))\in (\Lambda_*)_+$, which shows (ii).

By (i) of lemma 3.3, $(\Gamma,\Gamma_+)$ is a dimension group, so it is weakly unperforated and satisfies the Riesz interpolation property, which coincides with the Riesz decomposition property (see [Reference Goodearl30, proposition 2.1]). The correspondence $H_0\oplus p(I)\mapsto H_1\oplus p(I)$ preserves finite sums and intersections. Moreover, no subquotient of Γ is liminary by (v) of lemma 3.3, so $\Lambda_*$ satisfies the Riesz decomposition property by [Reference Elliott and Gong23, theorem 4.28], thus proving (iii).

Condition (iv) follows from [Reference Elliott20, theorem 5.2] and the fact that $H_1\oplus G$ is torsion free.

Proof. Consider the graded group $\Lambda_*=\Gamma\oplus (H_1\oplus G)$ as in lemma 3.5. Combining (iii) and (iv) of lemma 3.5 with [Reference Elliott21, theorem 8.1], it follows that there exists a separable stable A$\mathbb{T}$-algebra B with real rank zero such that (i) holds and $K_*(B)_+\cong (\Lambda_*)_+$. Since B is stable, we can identify $K_*(B)_+$ with the graded dimension range $D_*(B)$ (see §2.4), so (ii) is also satisfied.

We will prove that B is $\mathcal{Z}$-stable. Since B is an A$\mathbb{T}$-algebra, it follows that B has finite decomposition rank (see [Reference Kirchberg and Winter38, example 4.2]). We are going to use [Reference Robert and Tikuisis46, corollary 7.11]. Since B has finite decomposition rank and real rank zero, no quotient of B has a simple purely infinite ideal and the space of primitive ideals of B has a basis of compact open sets (see the remark (a) after [Reference Robert and Tikuisis46, corollary 7.11]). Moreover, the Murray-von Neumann semigroup of B is weakly divisible by (ii) of lemma 3.3, which yields that B has no nonzero elementary ideal quotients (see [Reference Thiel and Vilalta52, theorem 9.1] together with the implication (1)$\implies$ (3) of [Reference Thiel and Vilalta52, theorem B]). Therefore, B is $\mathcal{Z}$-stable by [Reference Robert and Tikuisis46, corollary 7.11].

Proof. Consider the automorphism of $\Gamma\oplus (H_1\oplus G)$ given by $(\kappa_0\oplus\alpha)\oplus(\kappa_1\oplus\alpha)$. Since $\alpha(G_+)=G_+$ by (iii) of lemma 3.1, $(\kappa_0\oplus\alpha)\oplus(\kappa_1\oplus\alpha)$ preserves $(\Lambda_*)_+$. Therefore, we can apply Elliott’s classification of automorphisms of A$\mathbb{T}$-algebras with real rank zero. Thus, by theorem 2.13, there exists $\gamma'\in \mathrm{Aut}(B)$ such that $K_0(\gamma')=\kappa_0\oplus\alpha$ and $K_1(\gamma')=\kappa_1\oplus\alpha$.

We claim that we can replace $\gamma'$ by another automorphism γ which induces the same maps in K-theory and the crossed product has finite nuclear dimension. Building on the work in [Reference Szabó, Wu and Zacharias50], we can take an automorphism γ of B with finite Rokhlin dimension such that $K_i(\gamma)=K_i(\gamma')$ for $i=0,1$ ([Reference Jacelon35, lemma 4.7]). Since B has finite nuclear dimension and γ has finite Rokhlin dimension, it follows that the crossed product $B\rtimes_{\gamma}\mathbb{Z}$ has finite nuclear dimension ([Reference Szabó, Wu and Zacharias50, theorem 6.2] or [Reference Hirshberg and Phillips33, theorem 3.1]).

We now show that $B\rtimes_{\gamma}\mathbb{Z}$ is simple. Since α ^k is non-trivial for all k ≠ 0, then $K_0(\gamma)^k$ is non-trivial, which implies that no non-trivial power of γ is inner. Moreover, since B is stable, the fact that $K_0(\gamma)^k$ is non-trivial, implies that there exists a projection $p_k\in B$ such that $\gamma^k(p_k)$ is not equivalent to p_k. Then, if u is a unitary in $\mathcal{M}(B)$, $\gamma^k(p_k)$ is not equivalent to $up_ku^*$, so

\begin{align*}\|\gamma^k(p_k)-up_ku^*\|=1.\end{align*}

By (iv) of lemma 3.3, the only order ideals I in Γ such that $(\kappa_0\oplus\alpha)(I)=I$ are $I=\{0\}$ and $I=\Gamma$. Thus, γ ^k is properly outer by the implication (iii)$\implies$(ii) in [Reference Olesen and Pedersen42, theorem 6.6]. Moreover, by (iv) of lemma 3.3, $K_0(B)$ has no non-trivial $K_0(\gamma)$-invariant ideals. Since B is an A$\mathbb{T}$-algebra of real rank zero, this implies that B has no non-trivial γ-invariant ideals ([Reference Rørdam48, proposition 1.5.3]), and hence the crossed product $B\rtimes_\gamma \mathbb{Z}$ is simple by [Reference Olesen and Pedersen42, theorem 7.2]. Furthermore, $B\rtimes_\gamma \mathbb{Z}$ has finite nuclear dimension, so it is $\mathcal{Z}$-stable by [Reference Tikuisis56, corollary 8.7].

Note that B has real rank zero, so it has an approximate unit of projections. Moreover, γ has finite Rokhlin dimension. Hence, (iv) follows from lemma 2.17.

Proof. Recall that $v=(w,1)\in \Gamma_+$, where $w\in H_0$ is such that $q_0(w)=[1_A]_0$. Let e be a projection in B such that $[e]_0=v$ in $K_0(B)$. We will prove that $A\cong e\left(B\rtimes_\gamma\mathbb{Z}\right)e$ using the Kirchberg–Phillips classification theorem ([Reference Kirchberg37, Reference Phillips44]). First, we claim that $e\left(B\rtimes_\gamma\mathbb{Z}\right)e$ is a simple, separable, unital, nuclear, purely infinite $\mathrm{C}^*$-algebra satisfying the $\mathrm{UCT}$.

Note that $B\rtimes_\gamma\mathbb{Z}$ is simple by (ii) of lemma 3.7, so e is a full projection. Moreover, B is nuclear, satisfies the UCT, and $B\rtimes_\gamma\mathbb{Z}$ is $\mathcal{Z}$-stable by (iii) of lemma 3.7. Therefore, by proposition 2.11$, e\left(B\rtimes_\gamma\mathbb{Z}\right)e$ is a simple, separable, unital, nuclear, $\mathcal{Z}$-stable $\mathrm{C}^*$-algebra which satisfies the UCT. By [Reference Rørdam49, corollary 5.1], to show that $e\left(B\rtimes_\gamma\mathbb{Z}\right)e$ is purely infinite, it now suffices to prove that $e\left(B\rtimes_\gamma\mathbb{Z}\right)e$ has no quasitraces. Since $e\left(B\rtimes_\gamma\mathbb{Z}\right)e$ is unital and nuclear, any quasitrace is a trace by [Reference Haagerup32]. Therefore, it suffices to show that $e\left(B\rtimes_\gamma\mathbb{Z}\right)e$ has no tracial states. If τ is a tracial state on $e\left(B\rtimes_\gamma\mathbb{Z}\right)e$, then it extends to a lower semicontinuous (possibly unbounded) trace $\tau'$ on $B\rtimes_\gamma\mathbb{Z}$ by [Reference Winter and Zacharias61, corollary 5.2]. By (iv) of lemma 3.7, the restriction of $\tau'$ to B induces a positive group homomorphism $\tau'_*:K_0(B)\to \mathbb{R}$ such that $\tau'_*(v)=1$ and $\tau'_*\circ K_0(\gamma)=\tau'_*$. Recall from (i) of lemma 3.6 and (i) of lemma 3.7 that $K_0(B)=\Gamma$ and $K_0(\gamma)=\kappa_0\oplus\alpha$. Thus, we obtain a contradiction with lemma 3.4. Hence, $e\left(B\rtimes_\gamma\mathbb{Z}\right)e$ has no tracial states. Moreover, it is nuclear and $\mathcal{Z}$-stable, so it is purely infinite by [Reference Rørdam49, corollary 5.1].

Therefore, by the Kirchberg–Phillips’ classification theorem, it suffices to check that $e\left(B\rtimes_\gamma\mathbb{Z}\right)e$ and A have the same Elliott invariant. First note that $e\left(B\rtimes_\gamma\mathbb{Z}\right)e$ and $B\rtimes_\gamma\mathbb{Z}$ have the same K-groups. Applying the Pimsner–Voiculescu exact sequence ([Reference Pimsner and Voiculescu45, theorem 2.4]) to B and the automorphism γ, we get the six-term exact sequence

By (i) of lemma 3.7, $\mathrm{id}_{H_1\oplus G}-K_1(\gamma)=(\mathrm{id}_{H_1}-\kappa_1)\oplus(\mathrm{id}_G-\alpha)$, which is injective by proposition 3.2 and since $\mathrm{id}_G-\alpha$ is an automorphism of G ((iii) of lemma 3.1). Then, the downward map $K_0(B\rtimes_\gamma\mathbb{Z})\to H_1\oplus G$ is zero. This further implies that the map $H_0\oplus G\to K_0(B\rtimes_\gamma\mathbb{Z})$ is surjective, which yields that

\begin{align*}K_0(B\rtimes_\gamma\mathbb{Z})\cong(H_0\oplus G)/(\mathrm{id}_{H_0\oplus G}-K_0(\gamma))(H_0\oplus G).\end{align*}

By lemma (i) of 2.7, $\mathrm{id}_{H_0\oplus G}-K_0(\gamma)=(\mathrm{id}_{H_0}-\kappa_0)\oplus(\mathrm{id}_G-\alpha)$. By (iii) of lemma 3.1, $\mathrm{id}_G-\alpha$ is an automorphism of G, so

\begin{align*}K_0(B\rtimes_\gamma\mathbb{Z})\cong H_0/(\mathrm{id}_{H_0}-\kappa_0)(H_0)\cong K_0(A),\end{align*}

where the last isomorphism is given by proposition 3.2. Note that $[e]_0=(w,1)$ in $K_0(B)$, where w is mapped to $[1_A]_0$ by the isomorphism $H_0/(\mathrm{id}_{H_0}-\kappa_0)(H_0)\cong K_0(A)$. Therefore, the resulting isomorphism $K_0(e\left(B\rtimes_\gamma\mathbb{Z}\right)e)\to K_0(A)$ sends $[e]_0$ to $[1_A]_0$.

Further examining the Pimsner–Voiculescu exact sequence and using that $\mathrm{id}_{H_0\oplus G}-K_0(\gamma)$ is injective, it follows that the map $K_1(B\rtimes_\gamma\mathbb{Z})\to H_0\oplus G$ is zero. Therefore, we obtain the exact sequence

which yields that $K_1(B\rtimes_\gamma\mathbb{Z})\cong (H_1\oplus G)/(\mathrm{id}_{H_1\oplus G}-(\kappa_1\oplus\alpha))(H_1\oplus G)\cong H_1/(\mathrm{id}_{H_1\oplus G}-\kappa_1)(H_1)$, where the last identification follows as $\mathrm{id}_G-\alpha$ is an automorphism on G ((iii) of lemma 3.1). Hence $K_1(B\rtimes_\gamma\mathbb{Z})\cong K_1(A)$ by proposition 3.2. Thus, $e\left(B\rtimes_\gamma\mathbb{Z}\right)e\cong A$ by the Kirchberg–Phillips’ classification theorem (see for example [Reference Rørdam48, theorem 8.4.1(iv)]).

Proof. We will only comment on the proof of the last statement. Let ω be a β-$\mathrm{KMS}$ state for the dual action $\hat{\rho}$ on $q(D \rtimes_\rho \mathbb{Z})q$. Then, by [Reference Laca and Neshveyev40, remark 3.3], there exists a unique β-$\mathrm{KMS}$ weight $\hat{\omega}$ for the dual action on $D\rtimes_\rho\mathbb{Z}$ which extends ω. The result now follows by [Reference Thomsen54, lemma 3.1].

Proof of theorem A

The proof follows the strategy in [Reference Elliott, Sato and Thomsen26, lemma 5.8]. Recall that if $S=\emptyset$, then we take θ to be the trivial flow on A. Therefore, we assume that S is nonempty and we claim that the $\mathrm{KMS}$ bundle $(S^\theta,\pi^\theta)$ of θ is isomorphic to $(S,\pi)$. This will finish the proof.

Let $(\omega,\beta)\in S^\theta$. By [Reference Laca and Neshveyev40, remark 3.3], there exists a unique β-$\mathrm{KMS}$ weight $\hat{\omega}$ for the dual action $\hat{\gamma}$ on $B\rtimes_\gamma\mathbb{Z}$ which extends ω. Furthermore, $B\rtimes_\gamma\mathbb{Z}$ is simple by (ii) of lemma 3.7, so e is full in $B\rtimes_\gamma\mathbb{Z}$. Then, by proposition 3.9, $\hat{\omega}|_B$ is a densely defined lower semicontinuous trace on B such that $\hat{\omega}|_B\circ\gamma=e^{-\beta}\hat{\omega}|_B$ and $\hat{\omega}|_B(e)=1$. Therefore, $(\hat{\omega}|_B)_*$ is a positive homomorphism on $K_0(B)$ such that

\begin{align*}(\hat{\omega}|_B)_*\circ K_0(\gamma)=e^{-\beta}(\hat{\omega}|_B)_* \ \text{and}\ (\hat{\omega}|_B)_*([e]_0)=1.\end{align*}

Then, by lemma 3.4, there exists a unique $\omega'\in\pi^{-1}(\beta)$ such that

(2.1)\begin{align} (\hat{\omega}|_B)_*(h,g)=g(\omega'), \end{align}

for all $(h,g)\in\Gamma\cong K_0(B)$. Define $\varphi:S^{\theta}\to S$ by $\varphi(\omega,\beta)=\omega'$ and note that $\pi\circ\varphi=\pi^\theta$ as $\omega'\in \pi^{-1}(\beta)$. Moreover, the restriction $\varphi:(\pi^\theta)^{-1}(\beta)\to \pi^{-1}(\beta)$ is affine by construction for any $\beta\in\mathbb{R}$.

We claim that φ is surjective. Let $\mu\in S$ and define $\mathrm{ev}_{\mu}:K_0(B)\to\mathbb{R}$ by $\mathrm{ev}_{\mu}(h,g)=g(\mu)$. Recall that $K_0(\gamma)=\kappa_0\oplus\alpha$ by (i) of lemma 3.7 and $\alpha(g)(x)=e^{-\pi(x)}g(x)$ for any $g\in G$. Then, $\mathrm{ev}_{\mu}$ is a positive group homomorphism such that $\mathrm{ev}_{\mu}\circ K_0(\gamma)=e^{-\pi(\mu)}\mathrm{ev}_{\mu}$ and $\mathrm{ev}_{\mu}(v)=1$ as $v=(w,1)$. Then, by proposition 2.15, there exists a unique densely defined lower semicontinuous trace τ_µ on B such that $(\tau_\mu)_*=\mathrm{ev}_\mu$. By proposition 3.9, there exists a $\pi(\mu)$-$\mathrm{KMS}$ state ω for θ such that $(\hat{\omega}|_B)_*=\mathrm{ev}_{\mu}$. Then $\varphi(\omega,\pi(\mu))=\mu$, so φ is surjective.

We further check that φ is injective. Let $(\omega_1,\beta_1),(\omega_2,\beta_2)\in S^\theta$ such that $\varphi(\omega_1,\beta_1)=\varphi(\omega_2,\beta_2)$. Then,

\begin{align*}\beta_1=\pi(\varphi(\omega_1,\beta_1))=\pi(\varphi(\omega_2,\beta_2))=\beta_2.\end{align*}

Moreover, by the definition of the map φ,

\begin{align*}(\hat{\omega}_1|_B)_*=(\hat{\omega}_2|_B)_*.\end{align*}

By proposition 3.9, $\hat{\omega}_1|_B$ and $\hat{\omega}_2|_B$ are densely defined lower semicontinuous traces on B such that

\begin{align*}(\hat{\omega}_1|_B)(e)=(\hat{\omega}_2|_B)(e)=1.\end{align*}

By proposition 2.15, it follows that $\hat{\omega}_1|_B=\hat{\omega}_2|_B.$ Then, proposition 3.9 yields that $\omega_1=\omega_2$, which shows that φ is injective.

If we show that $\varphi^{-1}:S\to S^\theta$ is continuous, then φ is a homeomorphism by lemma 2.5. Recall from remark 2.4 that both S ^θ and S are metrisable and let $\omega_n'$ be a sequence in S which converges to $\omega'$. If $\varphi^{-1}(\omega_n')=(\omega_n,\beta_n)$ and $\varphi^{-1}(\omega')=(\omega,\beta)$, then (2.1) yields that $(\hat{\omega}_n|_B)_*(h,g)$ converges to $(\hat{\omega}|_B)_*(h,g)$ for any $(h,g)\in \Gamma\cong K_0(B)$. Then, $\hat{\omega}_n|_B$ converges to $\hat{\omega}|_B$ by proposition 2.15, so ω_n converges to ω by proposition 3.9. Moreover, $\beta_n=\pi(\omega_n')$, which converges to $\pi(\omega')=\beta$ by continuity of π. Thus, $(\omega_n,\beta_n)$ converges to $(\omega,\beta)$, so φ ⁻¹ is continuous. Together with lemma 3.8, this yields the conclusion of theorem A.

Proof. Since S is compact and the topology on S is second countable, we can choose a countable subgroup G ₀ of $\mathcal{A}_0(S,\pi)$ such that for all ϵ > 0 and all $f\in \mathcal{A}_0(S,\pi)$, there is an element $g\in G_0$ such that

\begin{align*}\sup\limits_{x\in S}|f(x)-g(x)| \lt \epsilon.\end{align*}

Moreover, we can ensure that if $f\in G_0$, then

\begin{align*}x\mapsto e^{n\pi(x)}(1-e^{-\pi(x)})^mf(x)\end{align*}

is in G ₀ for all $m,n\in\mathbb{Z}$. Then, a similar construction to the one in [Reference Elliott and Thomsen27, §4.2 pp 109-110] allows to choose G ₀ such that both η and $\mathrm{id}-\eta$ are surjective, hence giving (ii). To check (iii), let $g\in G_0$ and $g_1,g_2\in\mathcal{A}_0(S,\pi)_+$ such that $g=g_1-g_2$. Using compactness of S, we can then ensure that $g_1,g_2\in G_0$.

Proof. Set

\begin{align*}G=\Bigg(\bigoplus\limits_{\mathbb{Z}}K_0(A)\Bigg)\oplus G_0.\end{align*}

We will define an order on G adapting the construction in [Reference Elliott and Thomsen27, §4.2]. The map

\begin{align*}r:\mathcal{A}(S,\pi)\to \mathrm{Aff}(T(A))\end{align*}

given by restriction is surjective ([Reference Elliott and Thomsen27, lemma 4.4(1)]), so we can choose a positive linear map

(3.1)\begin{align} L:\mathrm{Aff}(T(A))\to \mathcal{A}(S,\pi) \ \text{such that} \ r\circ L=\mathrm{id} \ \text{and} \ L(1)=1. \end{align}

If $\rho_A:K_0(A)\to \mathrm{Aff}(T(A))$ is the canonical pairing map of A, let us consider the homomorphism $\hat{L}:G\to \mathcal{A}(S,\pi)$ given by

(3.2)\begin{align} \hat{L}(\xi,g)(x)=g(x)+\sum\limits_{n\in\mathbb{Z}} L(\rho_A(\xi_n))(x) e^{n\pi(x)} \end{align}

for all $\xi=(\xi_n)_{n\in\mathbb{Z}}$, $g\in G_0$, and $x\in S$. Note that the map $\hat{L}$ is well-defined since ξ has only finitely many nonzero entries.

Define

(3.3)\begin{align} G_+=\{(\xi,g)\in G: \hat{L}(\xi,g)(x) \gt 0,\ x\in S\}\cup \{0\} \end{align}

and set $v=(1^{(0)},0),$ where $1^{(0)}$ is the sequence with $[1_A]_0$ in the zero entry and zero elsewhere.

To show (i), we will first show that the triple $(G,G_+,v)$ defines an ordered abelian group. The construction of $G_+$ in (3.3) yields that $G_+\cap(-G_+)=\{0\}$. We will check that $G=G_+-G_+$. Let $(\xi,g)\in G$. As A is stably finite, $(K_0(A),K_0(A)_+)$ is an ordered abelian group ([Reference Blackadar4, proposition 6.3.3]), so there exist sequences $\xi^{(1)},\xi^{(2)}$ such that $\xi^{(i)}_n\in K_0(A)_+,$ for any $i=1,2$ and $n\in\mathbb{N}$ and $\xi=\xi^{(1)}-\xi^{(2)}$. Moreover, by (iii) of lemma 4.1, there exist $g_1,g_2\in G_0\cap\mathcal{A}_0(S,\pi)_+$ such that $g=g_1-g_2$. Thus, $(\xi,g)=(\xi^{(1)},g_1)-(\xi^{(2)},g_2)$, which shows that $G=G_+-G_+$.

To show that $(G,G_+,v)$ is simple, we use the assumption that S is compact. Let $(\xi,g)\in G_+\setminus\{0\}$. By definition, this yields that $\hat{L}(\xi,g)(x) \gt 0$ for any $x\in S$. Since S is compact, there exists $n\in\mathbb{N}$ such that $\hat{L}(\xi,g)(x)\geq \frac{1}{n}$ for any $x\in S$. This implies that $(\xi,g)$ is an order unit, so $(G,G_+,v)$ is a simple ordered abelian group.

To see that $(G,G_+)$ is weakly unperforated, let $n\in\mathbb{N}$ and $(\xi,g)\in G$ such that $n(\xi,g)\in G_+\setminus\{0\}$. Therefore, by linearity of L, it follows that $\hat{L}(\xi,g)(x) \gt 0,\ x\in S$. Hence, $(\xi,g)\in G_+\setminus\{0\}$, so $(G,G_+)$ is weakly unperforated.

We claim that (ii) follows by combining (i) of lemma 4.1 and the fact that A has real rank zero. Let $f\in\mathcal{A}(S,\pi)$ and ϵ > 0. Note $f|_{T(A)}\in \mathrm{Aff}(T(A))$. Since A is unital, simple, exact, finite, $\mathcal{Z}$-stable, and has real rank zero, it follows that $\rho_A(K_0(A))$ is uniformly dense in $\mathrm{Aff}(T(A))$ ([Reference Rørdam49, theorem 7.2]). Therefore, there exists $y\in K_0(A)$ such that

\begin{align*} f(\tau)-\epsilon/2 \lt \rho_A(y)(\tau) \lt f(\tau)+\epsilon/2 \end{align*}

for any $\tau\in T(A)$. As T(A) is closed in S and S is compact, [Reference Bratteli, Elliott and Kishimoto9, lemma 2.3] shows that there exists $h\in\mathcal{A}(S,\pi)$ extending $\rho_A(y)$ such that

(3.4)\begin{align} f(x)-\epsilon/2 \lt h(x) \lt f(x)+\epsilon/2 \end{align}

for any $x\in S$. Then, $h-L(\rho_A(y))\in \mathcal{A}_0(S,\pi)$, so by (i) of lemma 4.1, there exists $g\in G_0$ such that

(3.5)\begin{align} |h(x)-L(\rho_A(y))(x)-g(x)| \lt \epsilon/2 \end{align}

for any $x\in S$. Now consider the element $(y^{(0)},g)\in G$, where $y^{(0)}$ is the sequence which is constant 0 apart from the zero entry which is equal to y. Then,

\begin{align*} \begin{array}{rcl} \sup\limits_{x\in S}|\hat{L}(y^{(0)},g)(x)-f(x)|&=&\sup\limits_{x\in S}|g(x)+L(\rho_A(y))(x)-f(x)|\\ &\stackrel{(3.5)}{\leq}&\sup\limits_{x\in S}|f(x)-h(x)|+\epsilon/2\\ &\stackrel{(3.4)}{\leq}& \epsilon. \end{array} \end{align*}

Thus, $\hat{L}(G)$ is uniformly dense in $\mathcal{A}(S,\pi)$.

We will now show (iii). This follows the strategy in (ii). Using [Reference Alfsen2, corollary II.3.11] and Kadison’s duality ([Reference Kadison36]), it suffices to check that G satisfies the strong Riesz interpolation property. Suppose that $(\xi_i,g_i) \lt (\eta_j,h_j)$ in G for any $i,j=1,2$. Therefore, $\hat{L}(\xi_i,g_i)(x) \lt \hat{L}(\eta_j,h_j)(x)$ for any $x\in S$. Moreover, $\hat{L}(\xi_i,g_i)$ and $\hat{L}(\eta_j,h_j)$ restrict to affine functions on T(A).

Since A is unital, simple, exact, finite, $\mathcal{Z}$-stable, and has real rank zero, it follows that $\rho_A(K_0(A))$ is uniformly dense in $\mathrm{Aff}(T(A))$ ([Reference Rørdam49, theorem 7.2]). Thus, $\rho_A(K_0(A))$ satisfies the strong Riesz interpolation property with respect to the strict ordering in $\mathrm{Aff}(T(A))$ ([Reference Effros, Handelman and Shen19, lemma 3.1]), so there exists $y\in K_0(A)$ such that

(3.6)\begin{align} \hat{L}(\xi_i,g_i)(\tau) \lt \rho_A(y)(\tau) \lt \hat{L}(\eta_j,h_j)(\tau) \end{align}

for any $\tau\in T(A)$ and any $i,j=1,2$. As T(A) is closed in S and S is compact, [Reference Bratteli, Elliott and Kishimoto9, lemma 2.3] shows that there exists $f\in\mathcal{A}(S,\pi)$ extending $\rho_A(y)$ such that

(3.7)\begin{align} \hat{L}(\xi_i,g_i)(x) \lt f(x) \lt \hat{L}(\eta_j,h_j)(x) \end{align}

for any $x\in S$ and any $i,j=1,2$.

Since S is compact, we can let δ > 0 be smaller than $f(x)-\hat{L}(\xi_i,g_i)(x)$ and $\hat{L}(\eta_j,h_j)(x)-f(x)$ for any $x\in S$ and any $i,j=1,2$. By (i) of lemma 4.1, there exists $g\in G_0$ such that

(3.8)\begin{align} |f(x)-L(\rho_A(y))(x)-g(x)| \lt \delta \end{align}

for any $x\in S$. Now consider the element $(y^{(0)},g)\in G$, where $y^{(0)}$ is the sequence which is constant 0 apart from the zero entry which is equal to y. We claim that

\begin{align*}\hat{L}(\xi_i,g_i)(x) \lt \hat{L}(y^{(0)},\quad g(x) \lt \hat{L}(\eta_j,h_j)(x)\end{align*}

for any $x\in S$. First note that for any $x\in S$ and $i=1,2$ we have that

\begin{align*} \begin{array}{rcl} \hat{L}(y^{(0)},g)(x)-\hat{L}(\xi_i,g_i)(x) &=&g(x)+L(\rho_A(y))(x)-\hat{L}(\xi_i,g_i)(x)\\ &\stackrel{(3.8)} { \gt }&f(x)-\delta-\hat{L}(\xi_i,g_i)(x)\\ & \gt &0, \end{array} \end{align*}

by the choice of δ. A similar calculation shows that $\hat{L}(\eta_j,h_j)(x) \gt \hat{L}(y^{(0)},g)(x)$ for any $x\in S$ and $j=1,2$. Hence, by the definition of the order on G, we obtain that

\begin{align*}(\xi_i,g_i) \lt (y^{(0)},g) \lt (\eta_j,h_j)\end{align*}

for any $i,j=1,2$, which shows that G satisfies the strong Riesz interpolation property. Thus, S(G) is a Choquet simplex. Moreover, S(G) is metrisable since G is countable.

We will now define the automorphism α of G. Let α be the automorphism of $(G,G_+,v)$ given by

\begin{align*}\alpha(\xi,f)=(\sigma(\xi),e^{-\pi}f)\end{align*}

where $\sigma(\xi)=(\xi_{n+1})_{n\in\mathbb{Z}}$ is the left shift on $\bigoplus\limits_{\mathbb{Z}}K_0(A)$ and e ^−π denotes the function $x\mapsto e^{-\pi(x)}$. To check (iv), let $(\xi,g)\in G$ such that $(\mathrm{id}-\alpha)(\xi,g)=0$. Then $\sigma(\xi)=\xi$ and $e^{-\pi}g=g$. Since σ is the shift on $\bigoplus_{\mathbb{Z}}K_0(A)$, it follows that ξ = 0. Moreover, g is supported away from $\pi^{-1}(0)$ by lemma 4.1, so g = 0. Thus, $\mathrm{id}-\alpha$ is injective.

We will now check (v). Consider the map $\Sigma_0:G\to K_0(A)$ given by

\begin{align*}\Sigma_0((\xi_n),f)=\sum\limits_{n\in\mathbb{Z}}\xi_n,\end{align*}

for any $(\xi_n)\in\bigoplus\limits_{\mathbb{Z}} K_0(A)$ and $f\in G_0$. We claim that the kernel of Σ₀ is $(\mathrm{id}-\alpha)(G)$. If $(\xi,g)\in G$, then $\Sigma_0((\mathrm{id}-\alpha)(\xi,g))=0$. Conversely, let $(\xi,g)\in G$ such that $\Sigma_0(\xi,g)=0$ and choose $N\in\mathbb{N}$ such that $\xi_n=0$ for all $|n|\geq N$. We follow the proof in [Reference Thomsen54, lemma 4.6]. Set $z_n=0$ for $n\geq N$ and

\begin{align*}z_n=\xi_n+z_{n+1},\ n \lt N.\end{align*}

Then

\begin{align*}z_{-N}=\xi_{-N}+\sum_{i=1}^{2N}\xi_{-N+i}=0\end{align*}

since $\Sigma_0(\xi,g)=0$. This yields that $z_n=0$ for any $n\leq -N$. Moreover, if we set $z=(z_n)_{n\in\mathbb{Z}}$, we get that $(\mathrm{id}-\sigma)(z)=\xi$. By condition (ii) in lemma 4.1, it also follows that there exists $f\in G_0$ such that $g=(1-e^{-\pi})f$, so $(\mathrm{id}-\alpha)(z,f)=(\xi,g)$. Thus, the kernel of Σ₀ is $(\mathrm{id}-\alpha)(G)$, so Σ₀ induces an isomorphism

\begin{align*}\Sigma: G/(\mathrm{id}-\alpha)(G)\to K_0(A).\end{align*}

The proof of (vi) follows as in [Reference Elliott, Sato and Thomsen26, lemma 3.8]. Let $\beta\in\mathbb R$ and φ be a state on $(G,G_+,v)$ such that $\varphi\circ\alpha=e^{-\beta}\varphi$.

The map $\hat{L}$ can be extended uniquely to a $\mathbb{Q}$-linear map on $\mathbb{Q}\otimes_{\mathbb{Z}}G$. For ease of notation, we will identify $\hat{L}$ with its extension $\hat{L}:\mathbb{Q}\otimes_{\mathbb{Z}}G\to \mathcal{A}(S,\pi)$. Then, we can also extend φ uniquely to a $\mathbb{Q}$-linear state $\hat{\varphi}$ of $\mathbb{Q}\otimes_{\mathbb{Z}}G$. Suppose that $\hat{L}(\xi,g)=0$ for some $(\xi,g)\in \mathbb{Q}\otimes_{\mathbb{Z}}G$. Then, $\frac{1}{n}v+(\xi,g),\frac{1}{n}v-(\xi,g)\in (\mathbb{Q}\otimes_{\mathbb{Z}}G)_+$, so

\begin{align*}-\frac{1}{n}\leq\hat{\varphi}(\xi,g)\leq \frac{1}{n}\end{align*}

for any $n\in\mathbb{N}$. Thus, $\hat{\varphi}(\xi,g)=0$, which implies that $\hat{\varphi}$ factors through $\hat{L}$. Indeed, there is a $\mathbb{Q}$-linear map

\begin{align*}\psi:\hat{L}(\mathbb{Q}\otimes_{\mathbb{Z}}G)\to\mathbb R, \ \text{such that}\ \psi\circ\hat{L}=\hat{\varphi}.\end{align*}

We claim that ψ is continuous. Let $f\in \hat{L}(\mathbb{Q}\otimes_{\mathbb{Z}}G)$ and suppose by compactness of S that $|f(x)| \lt \frac{n}{m}$ for any $x\in S$, for some $n,m\in\mathbb{N}$. Then, there exists $w\in\mathbb{Q}\otimes_{\mathbb{Z}}G$ such that $\hat{L}(w)=f$. Since

\begin{align*}-n=-n\hat{L}(v)(x) \lt m\hat{L}(w)(x) \lt n\hat{L}(v)(x)=n\end{align*}

for any $x\in S$, we get that $-nv \lt mw \lt nv$ in $\mathbb{Q}\otimes_{\mathbb{Z}}G$. Since $\psi\circ\hat{L}=\hat{\varphi}$ and $\hat{\varphi}(v)=1$, it follows that $|\psi(f)|=|\hat{\varphi}(w)|\leq\frac{n}{m}$. If there exists $g\in\hat{L}(\mathbb{Q}\otimes_{\mathbb{Z}}G)$ such that $|\psi(g)| \gt \sup\limits_{x\in S}|g(x)|$, then there exist $n,m\in\mathbb{N}$ such that

\begin{align*}|\psi(g)| \gt \frac{n}{m} \gt \sup\limits_{x\in S}|g(x)|.\end{align*}

But by the argument above, if $|g(x)| \lt \frac{n}{m}$ for any $x\in S$, then $|\psi(g)|\leq\frac{n}{m}$, so ψ must be norm-contractive.

It follows from (ii) above that $\hat{L}(\mathbb{Q}\otimes_{\mathbb{Z}}G)$ is uniformly dense in $\mathcal{A}(S,\pi)$. By continuity of ψ, we get that ψ extends uniquely to a linear norm-contractive map $\psi:\mathcal{A}(S,\pi)\to\mathbb{R}$. The fact that there exists a unique $s\in \pi^{-1}(\beta)$ such that $\psi=\mathrm{ev}_s$ follows as in [Reference Elliott, Sato and Thomsen26, lemma 3.8]. Hence, $\varphi=\mathrm{ev}_s\circ \hat{L}$ as required.

Proof. We apply [Reference Rørdam47, proposition 3.5] to the pair $(K_1(A),0)$.

Proof. To construct a $\mathrm{C}^*$-algebra B as in the statement of the lemma, we apply Elliott’s range-of-the-invariant result (see theorem 2.12). Lemma 4.2 shows that $(G,G_+,v)$ is a simple, weakly unperforated, countable ordered abelian group and S(G) is a metrisable Choquet simplex. Thus, by theorem 2.12, there exists a simple, separable, unital, nuclear, $\mathcal{Z}$-stable $\mathrm{C}^*$-algebra B, satisfying the $\mathrm{UCT}$ such that $\mathrm{Ell}(B)\cong (G,G_+,v, H, S(G), \rho_G).$

We construct an automorphism $\gamma'$ which satisfies (ii). Set $p_1=1_B\otimes e_{11}$, where e ₁₁ is a minimal projection in $\mathcal{K}$, and say that $\alpha([p_1]_0)=[p_2]_0$ for some projection $p_2\in B\otimes\mathcal{K}$. Then, essentially since any matrix amplification of $B\otimes\mathcal{K}$ is isomorphic to $B\otimes\mathcal{K}$, there exists a $^*$-isomorphism $\theta:B\otimes\mathcal{K}\to B\otimes\mathcal{K}\otimes\mathcal{K}$ such that $K_i(\theta)=K_i(\phi)$ for $i=0,1$ (see for example the argument in [Reference Wegge-Olsen59, corollary 6.2.11]), where $\phi:B\otimes\mathcal{K}\to B\otimes\mathcal{K}\otimes\mathcal{K}$ is given by $\phi(x)=x\otimes e_{11}$. By [Reference Brown10], we get isometries $v_i\in \mathcal{M}(B\otimes\mathcal{K}\otimes\mathcal{K})$ such that $v_iv_i^*=p_i\otimes 1_{\mathcal{M}(\mathcal{K})}$ for $i=1,2$. Thus, the maps

\begin{align*}\eta_i:=\mathrm{Ad}(v_i)\circ \phi: B\otimes\mathcal{K}\to p_i(B\otimes\mathcal{K})p_i\otimes\mathcal{K}\end{align*}

induce isomorphisms at the level of K ₀ and K ₁ for any $i=1,2$.

We then set

\begin{align*}\alpha':=K_0(\eta_2)\circ\alpha\circ K_0(\eta_1)^{-1}\end{align*}

to be a unital ordered group isomorphism from $K_0(p_1(B\otimes\mathcal{K})p_1)$ to $K_0(p_2(B\otimes\mathcal{K})p_2)$. Moreover, set

\begin{align*}\kappa':=K_1(\eta_2)\circ\kappa\circ K_1(\eta_1)^{-1}\end{align*}

to be a group isomorphism from $K_1(p_1(B\otimes\mathcal{K})p_1)$ to $K_1(p_2(B\otimes\mathcal{K})p_2)$.

By construction of B, the canonical map from tracial states on B to states on $K_0(B)$ is an affine homeomorphism. Therefore, the tracial states on $p_i(B\otimes\mathcal{K})p_i$ are determined by states on $K_0(p_i(B\otimes\mathcal{K})p_i)$ for any $i=1,2$. Hence, it is immediate to see that there exists a unique linear isomorphism $\sigma:\mathrm{Aff}(T(p_1(B\otimes\mathcal{K})p_1))\to \mathrm{Aff}(T(p_2(B\otimes\mathcal{K})p_2))$, which is compatible with $\alpha'$ via the canonical pairing maps. Moreover, since B is nuclear, $\mathcal{Z}$-stable, and satisfies the $\mathrm{UCT}$, and all these properties are preserved by stable isomorphisms (see proposition 2.10), so are $p_i(B\otimes\mathcal{K})p_i$ for $i=1,2$. Then, $p_1,p_2$ are full projections, so $p_i(B\otimes\mathcal{K})p_i$ is simple, separable, unital for $i=1,2$. Thus, by [Reference Carrión, Gabe, Schafhauser, Tikuisis and White13, corollary 9.5], there exists a unital $^*$-isomorphism $\psi:p_1(B\otimes\mathcal{K})p_1\to p_2(B\otimes\mathcal{K})p_2$ such that $(K_0(\psi), K_1(\psi), \mathrm{Aff}(T(\psi)))= (\alpha', \kappa', \sigma)$.

Set $\gamma'$ to be the following sequence of maps

By construction, we have that $K_0(\gamma')=\alpha$ and $K_1(\gamma')=\kappa$.

To check condition (iii), we will first show that we can choose γ such that the crossed product is simple and has finite nuclear dimension. Building on work in [Reference Szabó, Wu and Zacharias50], we can take an automorphism γ of $B\otimes\mathcal{K}$ with finite Rokhlin dimension such that $K_i(\gamma)=K_i(\gamma')$ for $i=0,1$ ([Reference Jacelon35, lemma 4.7] or [Reference Hirshberg, Winter and Zacharias34, theorem 3.4]). Since $B\otimes\mathcal{K}$ is simple, nuclear, and $\mathcal{Z}$-stable, it has finite nuclear dimension ([Reference Castillejos and Evington14, theorem A]). Moreover, γ has finite Rokhlin dimension, so the crossed product $(B\otimes\mathcal{K})\rtimes_{\gamma}\mathbb{Z}$ has finite nuclear dimension ([Reference Szabó, Wu and Zacharias50, theorem 5.2] or [Reference Hirshberg and Phillips33, theorem 3.1]). Since $K_0(\gamma)^n\neq \mathrm{id}$ for all n ≠ 0, no non-trivial power of γ is inner. Moreover, $B\otimes\mathcal{K}$ is simple, so $(B\otimes\mathcal{K})\rtimes_\gamma\mathbb{Z}$ is simple by [Reference Kishimoto39, theorem 3.1]. Thus, we get (iii) by [Reference Tikuisis56, corollary 8.7].

Since B is unital and $\mathcal{K}$ has real rank zero, $B\otimes\mathcal{K}$ has an approximate unit of projections. Moreover, γ has finite Rokhlin dimension, so (iv) follows from lemma 2.17.

Proof. We will use the classification theorem in [Reference Carrión, Gabe, Schafhauser, Tikuisis and White13, theorem 9.9]. First, we check that $p\left((B\otimes\mathcal{K})\rtimes_\gamma\mathbb{Z}\right)p$ is a simple, separable, unital, nuclear, $\mathcal{Z}$-stable $\mathrm{C}^*$-algebra satisfying the $\mathrm{UCT}$. Note that $(B\otimes\mathcal{K})\rtimes_\gamma\mathbb{Z}$ is simple by (iii) of lemma 4.5, so p is a full projection. Moreover, $B\otimes\mathcal{K}$ is nuclear, satisfies the UCT, and $(B\otimes\mathcal{K})\rtimes_\gamma\mathbb{Z}$ is $\mathcal{Z}$-stable by (iii) of lemma 4.5. Therefore, by proposition 2.11$, p\left((B\otimes\mathcal{K})\rtimes_\gamma\mathbb{Z}\right)p$ is a simple, separable, unital, nuclear, $\mathcal{Z}$-stable $\mathrm{C}^*$-algebra which satisfies the UCT.

It remains to check that the Elliott invariant of $p\left((B\otimes\mathcal{K})\rtimes_\gamma\mathbb{Z}\right)p$ is isomorphic to the Elliott invariant of A. As discussed in remark 2.9, we do not need to check that the positive cones in the K ₀-groups coincide. Note first that the space of tracial states on $p\left((B\otimes\mathcal{K})\rtimes_\gamma\mathbb{Z}\right)p$ is linearly isomorphic to the space of densely defined lower semicontinuous traces on $(B\otimes\mathcal{K})\rtimes_\gamma\mathbb{Z}$ by [Reference Cuntz and Pedersen18, proposition 4.7]. By (iv) of lemma 4.5, the latter is in a bijective correspondence to the space of γ-invariant densely defined lower semicontinuous traces on $B\otimes\mathcal{K}$. As observed previously, since B is simple, unital and $S(K_0(B))\cong T(B)$, it follows that we can identify the space of γ-invariant densely defined lower semicontinuous traces on $B\otimes\mathcal{K}$ with the space of $K_0(\gamma)$-invariant states on $K_0(B)$. But the latter is homeomorphic to $\pi^{-1}(0)\cong T(A)$ by (vi) of lemma 4.2. Hence, $T\left(p\left((B\otimes\mathcal{K})\rtimes_\gamma\mathbb{Z}\right)p\right)\cong T(A)$.

To compute the K-groups of $p\left((B\otimes\mathcal{K})\rtimes_\gamma\mathbb{Z}\right)p$, we apply the Pimsner–Voiculescu exact sequence in [Reference Pimsner and Voiculescu45, theorem 2.4] to the $\mathrm{C}^*$-algebra $B\otimes\mathcal{K}$ and the automorphism γ. Identifying $K_i(B\otimes\mathcal{K})$ with $K_i(B)$ for $i=0,1$, we get that

Recall from (i) of lemma 4.5 that $K_1(B)=H$ and from (ii) of lemma 4.5 that $K_1(\gamma)=\kappa$. Then, proposition 4.4 gives that $\mathrm{id}-K_1(\gamma)$ is injective. Therefore, the map $K_0((B\otimes\mathcal{K})\rtimes_\gamma\mathbb{Z})\to K_1(B)$ is zero, which yields that the map $K_0(B)\to K_0((B\otimes\mathcal{K})\rtimes_\gamma\mathbb{Z})$ is surjective. Thus,

(3.9)\begin{align} K_0((B\otimes\mathcal{K})\rtimes_\gamma\mathbb{Z})\cong K_0(B)/(\mathrm{id}-K_0(\gamma))(K_0(B)). \end{align}

Since $K_0(B)=G$ by (i) of lemma 4.5 and $K_0(\gamma)=\alpha$ by (ii) of lemma 4.5, we get that $K_0((B\otimes\mathcal{K})\rtimes_\gamma\mathbb{Z})\cong K_0(A)$ by (v) of lemma 4.2. This gives that,

\begin{align*}K_0\left(p\left((B\otimes\mathcal{K})\rtimes_\gamma\mathbb{Z}\right)p\right)\cong K_0(A).\end{align*}

Since $\Sigma([p]_0)=[1_A]_0$, it follows that the K ₀-isomorphism is compatible with the position of the unit.

Combining (ii) of lemma 4.5 and (iv) of lemma 4.2 yields that the map $\mathrm{id}-K_0(\gamma)$ is injective. Therefore, we obtain the short exact sequence

It follows that $K_1((B\otimes\mathcal{K})\rtimes_\gamma\mathbb{Z})\cong K_1(A)$ by proposition 4.4. Hence,

\begin{align*}K_1\left(p\left((B\otimes\mathcal{K})\rtimes_\gamma\mathbb{Z}\right)p\right)\cong K_1(A).\end{align*}

To apply [Reference Carrión, Gabe, Schafhauser, Tikuisis and White13, theorem 9.9], it remains to check that $p\left((B\otimes\mathcal{K})\rtimes_\gamma\mathbb{Z}\right)p$ and A have the same pairing between K-theory and traces. Let τ be a tracial state on A and $\tau_*$ be the induced state on $K_0(A)$. Recall that the homeomorphism from T(A) onto $T(p\left((B\otimes\mathcal{K})\rtimes_\gamma\mathbb{Z}\right)p)$ is given by following sequence of mappings. One first uses (vi) of lemma 4.2 to send τ to $\mathrm{ev}_{\tau}\circ \hat{L}$, which is a $K_0(\gamma)$-invariant state on $K_0(B)$. Then, (i) of lemma 4.5 yields that $\mathrm{ev}_{\tau}\circ \hat{L}$ corresponds to a unique γ-invariant tracial state on $B\otimes\mathcal{K}$. By (iv) of lemma 4.5, this extends uniquely to a densely defined lower semicontinuous trace on $(B\otimes\mathcal{K})\rtimes_\gamma\mathbb{Z}$. As mentioned previously, the latter corresponds to a tracial state on $p\left((B\otimes\mathcal{K})\rtimes_\gamma\mathbb{Z}\right)p$ ([Reference Cuntz and Pedersen18, proposition 4.7]).

Therefore, it suffices to check that

\begin{align*}\tau_*\circ\Sigma_0 = \mathrm{ev}_{\tau}\circ \hat{L},\end{align*}

where $\Sigma_0:K_0(B)\to K_0(A)$ is the homomorphism in (v) of lemma 4.2 which induces the isomorphism Σ. If $(\xi,g)\in K_0(B)$, then

\begin{align*}\tau_*\circ\Sigma_0(\xi,g)=\tau_*\left(\sum\limits_{n\in\mathbb{Z}}\xi_n\right)=\sum\limits_{n\in\mathbb{Z}}\rho_A(\xi_n)(\tau).\end{align*}

On the other hand,

\begin{align*}(\mathrm{ev}_{\tau}\circ\hat{L})(\xi,g)=g(\tau)+\sum\limits_{n\in\mathbb{Z}}L(\rho_A(\xi_n))(\tau)e^{n\pi(\tau)}.\end{align*}

Since $\pi(\tau)=0$ and $g\in G_0$ is supported away from T(A) by lemma 4.1, it follows that

\begin{align*}(\mathrm{ev}_{\tau}\circ\hat{L})(\xi,g)=\sum\limits_{n\in\mathbb{Z}}L(\rho_A(\xi_n))(\tau).\end{align*}

Furthermore, recall from (3.1) that for any $f\in \mathrm{Aff}(T(A))$, $L(f)|_{T(A)}=f$. Thus,

\begin{align*}(\mathrm{ev}_{\tau}\circ\hat{L})(\xi,g)=\sum\limits_{n\in\mathbb{Z}}\rho_A(\xi_n)(\tau)=\tau_*\circ\Sigma_0(\xi,g).\end{align*}

Hence, as A and $p\left((B\otimes\mathcal{K})\rtimes_\gamma\mathbb{Z}\right)p$ have the same Elliott invariant, they are isomorphic by [Reference Carrión, Gabe, Schafhauser, Tikuisis and White13, theorem 9.9].