Proof. (i) Assume that $\widehat {\mathcal {L}(X)}=\mathcal {L}(X)$ and let $x\in X$ . Let $N\in \mathbb {N}$ and set $v{:=}q x_{[-N,N]}$ . Since $\hat {v}\in \mathcal {L}(X)$ , there is some $y\in X$ such that $\hat {v}\prec y$ and by replacing y with $\varphi ^k(y)$ for some $k\in \mathbb {Z}$ if necessary, we can assume without loss of generality that $y_{[-N,N]}=\hat {v}$ . Since $\sigma (x)_{[-N,N]} = \hat {v}$ , we have that y approximates $\sigma (x)$ at a tolerance of $2^{-N}$ . As $N\in \mathbb {N}$ was arbitrary and X is closed, this shows that $\sigma (x)\in X$ . The reverse implication is immediate.

(ii) Suppose that $\mathcal {L}(X)$ contains anti-palindromes of arbitrarily large length, say $(w_n)_{n=1}^\infty $ with $|w_n|\nearrow \infty $ . Since anti-palindromes necessarily have even length, we write $|w_n|=2\ell _n$ for $n\ge 1$ . For each $n\ge 1$ , we thus obtain $x^{(n)} \in X$ such that $w_n\prec x^{(n)}$ and by shifting $x^{(n)}$ appropriately, we can assume without loss of generality that $x^{(n)}_{[-\ell _n+1,\ell _n]}=w_n$ . By compactness, the sequence $(x^{(n)})_{n=1}^\infty $ has a cluster point $x\in X$ and by passing to a subsequence if necessary, we can assume that $x^{(n)}\to x$ . Let $j\ge 1$ and pick $n\in \mathbb {N}$ large enough so that $\ell _n\ge j$ and $x_{[-j+1,j]}^{(n)}=x_{[-j+1,j]}$ . Since

$$ \begin{align*} w_n&=x_{-\ell_n+1}^{(n)}x_{-\ell_n+2}^{(n)}\cdots x_{-1}^{(n)}x_0^{(n)}x_1^{(n)}\cdots x_{\ell_n}^{(n)} \\ &= x_{-\ell_n+1}x_{-\ell_n+2}\cdots x_{-1}x_0x_1\cdots x_{\ell_n} \end{align*} $$

is an anti-palindrome, $x_i=\mathsf {1}-x_{-i+1}$ for all $-\ell _n+1\le i\le \ell _n$ and, in particular, $x_j=\mathsf {1}-x_{-j+1}$ . As $j\ge 1$ was arbitrary, we conclude that x is an anti-palindromic sequence.

The ‘only if’ part of the statement is immediate.

Finally, the last statement follows from [Reference Ortega and Scarparo37, Proposition 2.8].

Proof. (i) If $v\in \mathcal {L}(\underline {t})$ , then $v\prec t_n$ for some $n\in \mathbb {N}$ and so $\hat {v}\prec \hat {t}_n\prec t_{n+1}\prec \underline {t}$ .

(ii) We claim that for $p,n\in \mathbb {N}$ , if $t_n=x_1\cdots x_{2^{n+1}-1}$ ( $x_i\in \{\mathsf {0},\mathsf {1}\}$ ), then

(2.6) $$ \begin{align} t_{p+n+1} = t_p x_1 \hat{t}_p x_2 t_p x_3 \hat{t}_p\cdots x_{2^{n+1}-2} t_p x_{2^{n+1}-1}\hat{t}_p. \end{align} $$

Fix $p\in \mathbb {N}$ ; we shall prove the claim by induction on n. For $n=0$ , this is clear by definition, since $t_{p+1} = t_p\mathsf {1}\hat {t}_p$ (and $\mathsf {1}=t_0$ ). Assume that the claim has been proved for some $n\in \mathbb {N}$ , so

$$ \begin{align*} t_{p+n+1} = t_p x_1 \hat{t}_p x_2 t_p x_3 \hat{t}_p\cdots x_{2^{n+1}-2} t_p x_{2^{n+1}-1}\hat{t}_p \end{align*} $$

with $t_n = x_1\cdots x_{2^{n+1}-1}$ . Then, since $t_{p+n+2} = t_{p+n+1} \mathsf {1} \hat {t}_{p+n+1}$ ,

$$ \begin{align*} t_{p+n+2} &= t_px_1\hat{t}_px_2t_p\cdots t_px_{2^{n+1}-1}\hat{t}_p\mathsf{1} t_p \hat{x}_{2^{n+1}-1} \hat{t}_p \cdots \hat{t}_p \hat{x}_2 t_p \hat{x}_1 \hat{t}_p \end{align*} $$

and note that

$$ \begin{align*} t_{n+1}&=t_{n}\mathsf{1}\hat{t}_{n} \\ &=x_1x_2\cdots x_{2^{n+1}-1}\mathsf{1} \hat{x}_{2^{n+1}-1}\cdots\hat{x}_2\hat{x}_1 \end{align*} $$

as we wanted, and so (2.6) holds.

To see that $\underline {t}$ is uniformly recurrent, let $v\prec \underline {t}$ be a finite subword. By definition, there is $p\in \mathbb {N}$ such that $v\prec t_p$ . Let $w\prec \underline {t}$ be any subword with $|w|=3\cdot 2^{p+1}$ . Let $N\ge p+2$ be such that $w\prec t_N$ and write $N=p+n+1$ for some $n\ge 1$ . Note that by (2.6) and since $|w|=3\cdot 2^{p+1}$ , we must have that $t_p\prec w$ and thus $v\prec w$ as we wanted.

Assume now that $\underline {t}$ is periodic, that is, $\underline {t}=wvvv\cdots $ for some $w,v\in \{\mathsf {0,1}\}^*$ . By replacing v with $vv$ , we can assume without loss of generality that the length of v is even. Let $c,d\ge 0$ be such that $|v| = 2^{c+1}(2d+1)$ . By replacing v with some cyclic permutation of its letters if necessary (and thus without affecting its length), we can assume without loss of generality that $w=t_{cq}$ for some $q\ge 1$ . By (2.6), we have that

$$ \begin{align*} t_{cq+d+1} &= t_{c+ c(q-1)+d+1} \\ &=t_c x_1 \hat{t}_c \cdots t_c x_{2^{c(q-1)}-1}\hat{t}_c x_{2^{c(q-1)}} t_c x_{2^{c(q-1)}+1} \hat{t}_c \cdots t_c x_{2^{c(q-1)+d+1}-1}\hat{t}_c, \end{align*} $$

with $x_1\cdots x_{2^{c(q-1)+d+1}-1}=t_{c(q-1)+d}$ . Note that $t_{cq+d+1}$ is an initial segment of $\underline {t}$ and its initial segment

$$ \begin{align*} t_cx_1\hat{t}_c \cdots t_cx_{2^{c(q-1)}-1}\hat{t}_c \end{align*} $$

has length $2^{cq+1}-1$ , and so is equal to $t_{cq}=w$ . Therefore, the periodic part $vvv\cdots $ has $x_{2^{c(q-1)}} t_c x_{2^{c(q-1)}+1} \hat {t}_c \cdots t_c x_{2^{c(q-1)+d+1}-1}\hat {t}_c $ as an initial segment, which in turn has $x_{2^{c(q-1)}}t_cx_{2^{c(q-1)}+1} \hat {t}_cx_{2^{c(q-1)}+2}\cdots \hat {t}_cx_{2^{c(q-1)}+2d} t_c$ as an initial segment of length $2^{c+1}(2d+1)=|v|$ , so

$$ \begin{align*} v=x_{2^{c(q-1)}}t_cx_{2^{c(q-1)}+1} \hat{t}_cx_{2^{c(q-1)}+2}\cdots \hat{t}_cx_{2^{c(q-1)}+2d} t_c. \end{align*} $$

This is succeeded by $x_{2^{c(q-1)}+2d+1}\hat {t}_c$ , which is an initial segment of v, and so it follows that $t_c=\hat {t}_c$ . This is a contradiction, as this would imply that $|t_c|_{\mathsf {0}}=|t_c|_{\mathsf {1}}$ , which is never the case since $|t_c|_{\mathsf {1}}=|t_c|_{\mathsf {0}}+1$ by construction.

(iii) Note first that any anti-palindrome has even length. Now, if $v=v_1\cdots v_{2n}\in \{\mathsf {0},\mathsf {1}\}^*$ is an anti-palindrome of length $2n>8$ , then

$$ \begin{align*} v_{n-3}v_{n-2}v_{n-1}v_nv_{n+1}v_{n+2}v_{n+3}v_{n+4} \end{align*} $$

is an anti-palindrome of length $8$ that is a subword of v. It thus suffices to show that $\underline {t}$ contains no anti-palindromes of length $8$ , that is, that none of the $t_p$ contain anti-palindromes of length $8$ . We directly see that

$$ \begin{align*} t_4 = \mathsf{1101100111001001110110001100100} \end{align*} $$

contains no anti-palindromes of length $8$ . We continue with induction: assume we have shown that $t_p$ contains no anti-palindromes of length $8$ for some $p\ge 4$ . By (2.6), we have $t_{p} = t_3 s \hat {t}_3$ for some $s\prec \underline {t}$ (with $|s|= 2^{p+1}-2^5+1$ ) and $t_{p+1} = t_3 s \hat {t}_3 \mathsf {1} t_3 \hat {s} \hat {t}_3$ . Since $|t_3| = 2^4-1=15$ , if an anti-palindrome of length $8$ occurs as a subword of $t_{p+1}$ , then it is either a subword of $t_3s\hat {t}_3=t_p$ , or a subword of $\hat {t}_3\mathsf {1} t_3$ , or a subword of $t_3\hat {s}\hat {t}_3=\hat {t}_p$ . The first case and the third case are both impossible due to the inductive hypothesis (and upon noting that if $v\prec \hat {t}_p$ is an anti-palindrome of length $8$ , then $v=\hat {v}\prec t_p$ is an anti-palindrome of length $8$ ). As for the second case, we directly see that

$$ \begin{align*} \hat{t}_3\mathsf{1}t_3 = \mathsf{110110001100100\;1\;110110011100100} \end{align*} $$

contains no anti-palindromes of length $8$ and, thus, $t_{p+1}$ contains no anti-palindromes of length $8$ . The proof is complete by induction.

Proof. To define such an $\tilde {x}=(\tilde {x}_j)_{j\in \mathbb {Z}}$ , we first define $\tilde {x}_j$ for $j\ge 0$ such that $\tilde {x}_{[0,\infty )}=s$ . For $j\le -1$ , we define $\tilde {x}_j$ recursively so that $\tilde {x}_j\cdots \tilde {x}_{j+n}\in \mathcal {L}(\underline {s})$ for all $n\in \mathbb {N}$ . This suffices to conclude that $\mathcal {L}(\tilde {x})=\mathcal {L}(\underline {s})$ . Let $j\le -1$ and assume that $\tilde {x}_{j+1},\tilde {x}_{j+2},\ldots $ have all been defined so that

$$ \begin{align*} \tilde{x}_{j+1}\cdots \tilde{x}_{j+n}\in\mathcal{L}(\underline{s}) \end{align*} $$

for all $n\ge 1$ . We claim that there is an $a\in \mathsf {A}$ such that $a\tilde {x}_{j+1}\cdots \tilde {x}_{j+n}\in \mathcal {L}(\underline {s})$ for all $n\ge 1$ . Indeed, if this was not the case, then for any $a\in \mathsf {A}$ , there is some $n_a\ge 1$ such that $a\tilde {x}_{j+1}\cdots \tilde {x}_{j+n_a}\not \in \mathcal {L}(\underline {s})$ . Setting $n{:=}q \max \{n_a:a\in \mathsf {A}\}$ , we conclude that $a\tilde {x}_{j+1}\cdots \tilde {x}_{j+n}\not \in \mathcal {L}(\underline {s})$ for all $a\in \mathsf {A}$ . However, since $\underline {s}$ is (uniformly) recurrent and $\tilde {x}_{j+1}\cdots \tilde {x}_{j+n}$ is a subword of $\underline {s}$ , this finite subword occurs in $\underline {s}$ preceded by some letter $a\in \mathsf {A}$ , and so $a\tilde {x}_{j+1}\cdots \tilde {x}_{j+n}\in \mathcal {L}(\underline {s})$ , which is a contradiction.

It follows from the preceding paragraph that X is non-empty and it is actually infinite, since $\{\varphi ^k(\tilde {x})\}_{k\ge 0}$ is infinite, due to $\underline {s}$ being non-periodic. Note that $\varphi (X)=X$ since $\mathcal {L}(\varphi (x))=\mathcal {L}(x)$ for any $x\in \mathsf {A}^{\mathbb {Z}}$ . That X is closed follows from the fact that $\underline {s}$ is uniformly recurrent: let $(x^{(m)})_{m=1}^\infty \subset X$ be a sequence converging to $x\in \mathsf {A}^{\mathbb {Z}}$ , let $v\prec x$ and let $N\in \mathbb {N}$ be such that $v\prec x_{[-N,N]}$ . By taking m large enough, $x^{(m)}$ agrees with x on the slots $[-N,N]$ and so $v\prec x^{(m)}$ ; hence, $v\in \mathcal {L}(x^{(m)})=\mathcal {L}(\underline {s})$ , which shows that $\mathcal {L}(x)\subset \mathcal {L}(\underline {s})$ and so $x\in X$ .

Now, if $x\in X$ and $v\in \mathcal {L}(\underline {s})$ , since $\underline {s}$ is uniformly recurrent, there is $\ell \in \mathbb {N}$ such that if $w\prec \underline {s}$ with $|w|\ge \ell $ , then $v\prec w$ . Since $x_{[1,\ell ]}\in \mathcal {L}(x)\subset \mathcal {L}(\underline {s})$ has length $\ell $ , we get that $v\prec x_{[1,\ell ]}\prec x$ , and thus, $\mathcal {L}(\underline {s})\subset \mathcal {L}(x)$ . This shows that $X = \{ x\in \mathsf {A}^{\mathbb {Z}}: \mathcal {L}(x)=\mathcal {L}(\underline {s})\}$ , and thus X is minimal since $\mathcal {L}(x)=\mathcal {L}(X)$ for all $x\in X$ . In particular, X is a Cantor space as an infinite minimal subshift.

Proof. Note first that the substitution $\varrho $ is left-proper (cf. [Reference Durand, Host and Skau18, Definition 16]), in the sense that there is $p\in \mathbb {N}$ so that the words $\varrho ^p(a)$ start with the same letter for all $a\in \mathsf {A}$ ; in our case, one sees that for $p=2$ , all words $\varrho ^2(\mathsf {0})=\mathsf {3020},\varrho ^2(\mathsf {1})=\mathsf {3021},\varrho ^2(\mathsf {2})=\mathsf {3120}$ and $\varrho ^2(\mathsf {3})=\mathsf {3121}$ start with the letter $\mathsf {3}$ . We can thus apply [Reference Durand, Host and Skau18, Corollary 33] (together with [Reference Durand, Host and Skau18, §2.4.1, Remark(ii)]) and so $\mathrm {K}_0(C(X_\varrho )\rtimes _\varphi \mathbb {Z})$ is computed as an inductive limit: with M the matrix of (3.4), for $n\ge 1$ , consider the subgroups of $\mathbb {Q}^4$ given by

(3.6) $$ \begin{align} G_n {:=}q \{\vec{q}\in \mathbb{Q}^4: M^n\cdot \vec{q} \in \mathbb{Z}^4\}, \end{align} $$

(3.7) $$ \begin{align}\kern-5pt H_n{:=}q \{\vec{q}\in\mathbb{Q}^4: M^n\cdot\vec{q} = \vec{0}\} \end{align} $$

and the subsets

(3.8) $$ \begin{align} (G_{n})_+{:=}q\{\vec{q}\in \mathbb{Q}^4: M^n\cdot\vec{q}\in\mathbb{Z}_{\ge0}^4\}. \end{align} $$

Note that $G_n\subset G_{n+1}$ , $H_n\subset G_n$ , $H_n \subset H_{n+1}$ and that $(G_{n})_+\subset (G_{n+1})_+$ for all $n\ge 1$ . Set $G{:=}q \bigcup _{n\ge 1}G_n$ , $H{:=}q \bigcup _{n\ge 1}H_n \subset G$ and $G_+{:=}q \bigcup _{n\ge 1}(G_n)_+$ . Then, by [Reference Durand, Host and Skau18, §6.4 and Corollary 33], we have that $\mathrm {K}_0(C(X_\varrho )\rtimes _\varphi \mathbb {Z})$ is isomorphic as an abelian group to $G/H$ , its positive cone is identified with the image of $G_+$ under the quotient map and the order unit is identified with the equivalence class of $(1,1,1,1)$ in $G/H$ . By induction, one checks that

(3.9) $$ \begin{align} M^{n+2}=\begin{pmatrix} 2^n+1 & 2^n-1 & 2^n & 2^n \\ 2^n & 2^n & 2^n & 2^n \\ 2^n & 2^n & 2^n & 2^n \\ 2^n-1 & 2^n+1 & 2^n &2^n\end{pmatrix}, \quad n\in\mathbb{N}, \end{align} $$

and so with the notation $\vec {q}=(q_1,\ldots ,q_4)\in \mathbb {Q}^4$ , we obtain

(3.10) $$ \begin{align} G_{n+2}= \{\vec{q}\in\mathbb{Q}^4: 2^n(q_1+\cdots+q_4) \in \mathbb{Z},\; q_1-q_2\in\mathbb{Z}\}, \end{align} $$

(3.11) $$ \begin{align} H_{n+2} = \{\vec{q}\in\mathbb{Q}^4: q_1+\cdots+q_4=0,\; q_1=q_2\} \end{align} $$

and

(3.12) $$ \begin{align} (G_{n+2})_+ = \{ \vec{q}\in\mathbb{Q}^4: 2^n(q_1+\cdots+q_4)\in\mathbb{N},\;|q_1-q_2|\le2^n(q_1+\cdots+q_4)\}. \end{align} $$

We thus define group homomorphisms $\alpha _n\colon G_{n+2}\to {1}/{2^n}\mathbb {Z}\oplus \mathbb {Z}$ given by $\alpha _n(\vec {q})=(q_1+\cdots +q_4,q_1-q_2)$ . It is clear that each $\alpha _n$ is surjective, that $\ker (\alpha _n)=H_{n+2}$ and that

$$ \begin{align*} \alpha_n((G_{n+2})_+) = \bigg\{(s,m)\in\dfrac{1}{2^n}\mathbb{Z}\oplus \mathbb{Z}: s\ge 0, |m|\le 2^ns\bigg\}. \end{align*} $$

Denoting by $\tilde {\alpha }_n\colon G_{n+2}/H_{n+2}\to {1}/{2^n}\mathbb {Z}\oplus \mathbb {Z}$ the induced isomorphisms, we see that we have a commutative diagram:

(3.13)

where the arrows on each row denote the inclusion maps. The abelian group that is the inductive limit of the bottom row is $\mathbb {Z}[\tfrac {1}{2}]\oplus \mathbb {Z}$ , so we obtain an isomorphism $G/H\cong \mathbb {Z}[\tfrac {1}{2}]\oplus \mathbb {Z}$ . Moreover, the positive cone of $G/H$ (being the image of $G_+$ under the quotient map) is carried by this isomorphism to $\bigcup _{n\in \mathbb {N}}\tilde {\alpha }_n((G_{n+2})_+)$ , which is precisely the set

(3.14) $$ \begin{align} \bigcup_{n\in\mathbb{N}}\bigg\{(s,m)\in\dfrac{1}{2^n}\mathbb{Z}\oplus \mathbb{Z}: s\ge0, |m|\le 2^n s\bigg\} , \end{align} $$

which is in fact equal to $\{(s,m)\in \mathbb {Z}[\tfrac {1}{2}]\oplus \mathbb {Z}: s>0\}\cup \{(0,0)\}$ : the latter contains the set appearing in (3.14) which obviously contains $(0,0)$ . However, if $(s,m)\in \mathbb {Z}[\tfrac {1}{2}]\oplus \mathbb {Z}$ is such that $s>0$ , then we can write $s={k}/{2^n}$ for some $k,n\in \mathbb {N}$ . Now, take $n'\in \mathbb {N}$ large enough so that $|m|\le 2^{n'}k $ , and note that $s = ({2^{n'}k}/{2^{n+n'}}) \in ({1}/{2^{n+n'}})\mathbb {Z}$ and $2^{n+n'}s=2^{n'}k$ .

Finally, the isomorphism $G/H\cong \mathbb {Z}[\tfrac {1}{2}]\oplus \mathbb {Z}$ maps the equivalence class of $(1,1,1,1)$ to the element $(4,0)$ . This shows that, as a scaled ordered group, $\mathrm {K}_0(C(X_\varrho )\rtimes _\varphi \mathbb {Z})$ is isomorphic to

$$ \begin{align*} \big(\mathbb{Z}\big[\tfrac{1}{2}\big]\oplus\mathbb{Z}, \{(s,n): s>0,n\in\mathbb{Z}\}\cup\{(0,0)\},(4,0)\big), \end{align*} $$

which is, in turn, isomorphic to the scaled ordered group in (3.5), since $\mathbb {Z}[\tfrac {1}{2}]$ is $2$ -divisible.

Proof. For the first part of the statement, from Proposition 2.3, we obtain $y\in X_\varrho \subset \mathsf {A}^{\mathbb {Z}}$ such that $y_{[0,\infty )}=\underline {r}$ . Consider $\tilde {x}\in \{\mathsf {0},\mathsf {1}\}^{\mathbb {Z}}$ defined as $\tilde {x}_{[0,\infty )}=\underline {t}$ and, for $j\le -1$ , we choose $\tilde {x}_{2j},\tilde {x}_{2j+1}\in \{\mathsf {0},\mathsf {1}\}$ in the unique way so that $\mathsf {2}\tilde {x}_{2j}+\tilde {x}_{2j+1}=y_j$ . By definition of $\beta $ together with Remark 3.1, $\beta (\tilde {x})=y$ , so what we need to show is that $\tilde {x}\in \underline {X}$ . Indeed, let $k,\ell \in \mathbb {Z}$ with $k\le \ell $ and consider the subword $\tilde {x}_{2k}\tilde {x}_{2k+1}\cdots \tilde {x}_{2\ell }\tilde {x}_{2\ell +1}$ of $\tilde {x}$ . Since $\beta (\tilde {x})=y$ and $\mathcal {L}(y)=\mathcal {L}(\underline {r})$ , we have that $w{:=}q (\mathsf {2}\tilde {x}_{2k}+\tilde {x}_{2k+1})\cdots (\mathsf {2}\tilde {x}_{2\ell }+\tilde {x}_{2\ell +1})$ is a subword of $\underline {r}=(\mathsf {2}\tilde {x}_0+\tilde {x}_1)(\mathsf {2}\tilde {x}_2+\tilde {x}_3)\cdots $ , and so there exists $m\ge 0$ such that

$$ \begin{align*} w = (\mathsf{2}\tilde{x}_{2m}+\tilde{x}_{2m+1})\cdots(\mathsf{2}\tilde{x}_{2m+2\ell-2k} +\tilde{x}_{2m+2\ell-2k+1}). \end{align*} $$

Since $\mathsf {2}a+b=\mathsf {2}c+d$ implies that $a=b$ and $c=d$ for all $a,b,c,d\in \{\mathsf {0},\mathsf {1}\}$ , it follows that

$$ \begin{align*} \tilde{x}_{2k}\tilde{x}_{2k+1}\cdots\tilde{x}_{2\ell}\tilde{x}_{2\ell+1} = \tilde{x}_{2m}\tilde{x}_{2m+1}\cdots\tilde{x}_{2m+2\ell-2k}\tilde{x}_{2m+2\ell-2k+1}, \end{align*} $$

which is a subword of $\underline {t}=\tilde {x}_{[0,\infty )}$ as we wanted.

For the second part, item (i) is obvious. For item (ii), it is enough to show that $\underline {X}^{(0)} $ and $\underline {X}^{(1)}$ are disjoint since by minimality, their union is equal to $\underline {X}$ . To that end, observe first that $\underline {X}^{(0)} = \overline {\{\varphi _{\{\mathsf {0,1}\}}^{2k}(\tilde {x}): k\in \mathbb {N}\}}$ and that $\underline {X}^{(1)} = \overline {\{\varphi _{\{\mathsf {0,1}\}}^{2\ell +1}(\tilde {x}): \ell \in \mathbb {N}\}}$ (cf. [Reference Putnam40, Lemma 5.2]). Now, since $\tilde {x}_{[0,\infty )}=\underline {t}$ , it is enough to show that for any $k,\ell \in \mathbb {N}$ , we have that $\underline {t}_{[2k, 2k+6]} \ne \underline {t}_{[2\ell +1, 2\ell +7]}$ , since this implies that

(3.19) $$ \begin{align} d(\varphi_{\{\mathsf{0,1}\}}^{2k}(\tilde{x}), \varphi_{\{\mathsf{0,1}\}}^{2\ell+1}(\tilde{x}))\ge 2^{-7}, \quad k,\ell\in\mathbb{N}, \end{align} $$

where d is the metric defined in (2.1). Let $k,\ell \ge 0$ and let $n\in \mathbb {N}$ be such that $\max \{2k+6, 2\ell +7\}<2^{n+3}-1$ . With $t_{\mathrm {e}}{:=}q \underline {t}_{[2k,2k+6]}$ and $t_{\mathrm {o}}{:=}q \underline {t}_{[2\ell +1,2\ell +7]}$ , we have that $t_{\mathrm {e}}, t_{\mathrm {o}}\prec t_{n+2}$ . Applying (2.6) with $p=1$ and this n, we see that

(3.20) $$ \begin{align} t_{n+2} = \mathsf{110} x_1 \mathsf{100} x_2 \mathsf{110} x_3 \mathsf{100} x_4 \mathsf{110}x_5 \cdots \mathsf{110} x_{2^{n+1}-1}\mathsf{100} \end{align} $$

for some $x_i\in \{\mathsf {0,1}\}$ . Inspecting (3.20), we see that

(3.21) $$ \begin{align} t_{\mathrm{e}} \in \{ \mathsf{110}x\mathsf{100}, \mathsf{0}x\mathsf{100}y\mathsf{1}, \mathsf{100}x\mathsf{110}, \mathsf{0}x\mathsf{110}y\mathsf{1} \} \end{align} $$

for some $x,y\in \{\mathsf {0,1}\}$ , while

(3.22) $$ \begin{align} t_{\mathrm{o}}\in \{ \mathsf{10}x\mathsf{100}y, x\mathsf{100}y\mathsf{11}, \mathsf{00}x\mathsf{110}y, x\mathsf{110}y\mathsf{10}\} \end{align} $$

for some $x,y\in \{\mathsf {0,1}\}$ , and from this description, it is clear that $t_{\mathrm {e}}\ne t_{\mathrm {o}}$ .

For item (iii), note that by (3.16), we have that $\{\varphi _{\{\mathsf {0,1}\}}^{2k}(\tilde {x}):k\in \mathbb {Z}\}$ is mapped by $\beta $ onto the orbit of $\beta (\tilde {x})$ under the shift map on $\mathsf {A}^{\mathbb {Z}}$ . Since $\beta (\tilde {x})$ belongs to $X_\varrho $ which is a minimal subshift, $\beta (\underline {X}^{(0)})=X_\varrho $ .

Proof. Set $A{:=}q C(\underline {X})\rtimes _{\varphi _{\{\mathsf {0,1}\}}}\mathbb {Z}$ , $B{:=}q C(X_\varrho )\rtimes _{\varphi _{\mathsf {A}}}\mathbb {Z}$ and let $u_A\in A$ , $u_B\in B$ be the unitaries that are implementing the respective $\mathbb {Z}$ -actions, in the sense that $u_Afu_A^* = f\circ \varphi _{\{\mathsf {0,1}\}}^{-1}$ for all $f\in C(\underline {X})$ and $u_Bfu_B^* = f\circ \varphi _{\mathsf {A}}^{-1}$ for all $f\in C(X_\varrho )$ . We will show that $A\cong \mathrm {M}_2(B)$ , from which it follows that $\mathrm {K}_0(A) \cong \mathrm {K}_0(B)$ as ordered abelian groups, with $[1_A]_0$ identified with $2[1_B]_0$ , and so by Proposition 3.2 and $2$ -divisibility of $\mathbb {Z}[\tfrac {1}{2}]$ , it follows that $(\mathrm {K}_0(A),\mathrm {K}_0(A)_+,[1_A]_0)$ is isomorphic to

$$ \begin{align*} \big(\mathbb{Z}\big[\tfrac{1}{2}\big]\oplus\mathbb{Z}, \{(s,n): s>0,n\in\mathbb{Z}\}\cup\{(0,0)\},(1,0)\big). \end{align*} $$

Indeed, let $p\in C(\underline {X})$ be the indicator function of $\underline {X}^{(0)}$ and set $v{:=}q pu_A^*\in A$ . Note that $vv^* = p$ and that $v^*v = 1_A-p$ , since $\varphi _{\{\mathsf {0,1}\}}(\underline {X}^{(0)})=\underline {X}^{(1)}=\underline {X}\setminus \underline {X}^{(0)}$ . Sending v and $v^*$ to the matrix units $e_{12}$ and $e_{21}$ , respectively, $A\cong \mathrm {M}_2(pAp)$ and so it suffices to show that $B \cong pAp$ . Let $\vartheta \colon C(X_\varrho )\to C(\underline {X}^{(0)})\cong pC(\underline {X})p\subset pAp$ be the $^*$ -homomorphism

$$ \begin{align*} \vartheta(f) = f\circ\beta\vert_{\underline{X}^{(0)}},\quad f\in C(X_\varrho) \end{align*} $$

and let $\tilde {u}{:=}q pu_A^2p$ . Then,

$$ \begin{align*} \tilde{u}\vartheta(f)\tilde{u}^{\ast} &= f\circ \beta\vert_{\underline{X}^{(0)}}\circ \varphi_{\{\mathsf{0,1}\}}^{-2}\\ &\stackrel{{({3.16})}}{=}\; (f\circ\varphi_{\mathsf{A}}^{-1})\circ\beta\vert_{\underline{X}^{(0)}} \\ &= \vartheta (u_B fu_B^{\ast}), \end{align*} $$

and so we obtain a ${}^{\ast }$ -homomorphism $C(X_\varrho )\rtimes _{\varphi _{\mathsf {A}}}\mathbb {Z}\to pAp $ that is injective since the domain is simple [Reference Archbold and Spielberg2]. Moreover, it is surjective since the range clearly contains $pC(\underline {X})p$ and $\tilde {u}=pu_A^2p$ , while $pu_Ap=0$ .

Proof. We show that there is some $h\in C(\underline {X},\mathbb {Z})$ such that $h-\sigma _*(h)$ does not lie in $\mathrm {im}(1-\varphi _*)$ , which is equivalent to $[h] \ne \sigma _*[h]$ . Note that if $g\in \mathrm {im}(1-\varphi _*)$ , then by writing $g=f-f\circ \varphi ^{-1}$ for some $f\in C(\underline {X},\mathbb {Z})$ , we have that

(3.23) $$ \begin{align} \bigg|\!\sum_{j=0}^ng(\varphi^j(x))\bigg|&=|f(\varphi^n(x))-f(\varphi^{-1}(x))| \nonumber\\ &\le 2\|f\|_\infty \end{align} $$

for all $x\in \underline {X}$ and all $n\in \mathbb {N}$ , which is to say that

(3.24) $$ \begin{align} \sup_{x\in \underline{X},n\in\mathbb{N}}\bigg|\!\sum_{j=0}^ng(\varphi^j(x))\bigg|<\infty. \end{align} $$

(Note that (3.24) is also a sufficient condition for g to lie in $\mathrm {im}(1-\varphi _*)$ due to the Gottschalk–Hedlund theorem, but we do not make use of this.)

Let $[\mathsf {1}]$ and $[\mathsf {0}]$ denote the (clopen) subsets of $\underline {X}$ of those $(x_j)_{j\in \mathbb {Z}}\in \underline {X}$ such that $x_0=\mathsf {1}$ and $x_0=\mathsf {0}$ , respectively. Note that $\chi _{[\mathsf {1}]}-\sigma _*(\chi _{[\mathsf {1}]})=\chi _{[\mathsf {1}]}-\chi _{[\mathsf {0}]}$ and for the specific $g := \chi _{[\mathsf {1}]}-\chi _{[\mathsf {0}]} \in C(\underline {X},\mathbb {Z})$ , observe that for $x\in \underline {X}$ and $n\in \mathbb {N}$ , we have

(3.25) $$ \begin{align} \sum_{j=0}^ng(\varphi^j(x))=|x_{[0,n]}|_{\mathsf{1}}-|x_{[0,n]}|_{\mathsf{0}}. \end{align} $$

Let $\tilde {x}\in \underline {X}$ be such that $\tilde {x}_{[0,\infty )}=\underline {t}$ (cf. Proposition 2.3) and recursively define a sequence $(m_n)_{n=0}^\infty \subset \mathbb {N}$ by setting $m_0{:=}q 0$ and

(3.26) $$ \begin{align} m_{n+1}{:=}q\begin{cases} 2m_n+2\quad &\text{if } n \text{ is odd}, \\ 2m_n+1\quad &\text{if } n \text{ is even}. \end{cases} \end{align} $$

Note that $m_n$ is odd if and only if n is odd (and so $m_n$ is even if and only if n is even), and that $m_n\le 2^{n+1}-2$ for all $n\in \mathbb {N}$ . We claim that

(3.27) $$ \begin{align} |\underline{t}_{[0,m_n]} |_{\mathsf{1}} - |\underline{t}_{[0,m_n]}|_{\mathsf{0}} = n+1,\quad n\in\mathbb{N}, \end{align} $$

which we prove by induction. For $n=0$ , the claim is directly seen to be true. Assume now that the claim is true for some even $n\in \mathbb {N}$ and let us prove it for $n+1$ . Since $m_{n+1}\le 2^{n+2}-2$ and $|t_{n+1}|=2^{n+2}-1$ , we have that $\underline {t}_{[0,m_{n+1}]}$ is an initial segment of $t_{n+1}$ . By (2.6) (with $p=0$ therein),

$$ \begin{align*} t_{n+1} = \mathsf{1} x_1 \mathsf{0} x_2 \mathsf{1} x_3 \mathsf{0} x_4 \mathsf{1} \cdots \mathsf{0} x_{2^{n+1}-2} \mathsf{1} x_{2^{n+1}-1} \mathsf{0} \end{align*} $$

with $x_1\cdots x_{2^{n+1}-1}=t_{n}$ and so $x_1\cdots x_{m_n}x_{m_n+1} = \underline {t}_{[0,m_n]}$ . Since $m_{n+1}=2m_n+1$ and since $m_n$ is even,

$$ \begin{align*} \underline{t}_{[0,m_{n+1}]} = \mathsf{1} x_1 \mathsf{0} x_2 \mathsf{1}x_3 \mathsf{0}x_4\cdots \mathsf{0} x_{m_n-2} \mathsf{1} x_{m_n-1}\mathsf{0}x_{m_n}\mathsf{1}x_{m_n+1}, \end{align*} $$

whence

$$ \begin{align*} |\underline{t}_{[0,m_{n+1}]}|_{\mathsf{1}} = |\underline{t}_{[0,m_n]}|_{\mathsf{1}} + (m_n+1)/2, \end{align*} $$

while

$$ \begin{align*} |\underline{t}_{[0,m_{n+1}]}|_{\mathsf{0}} = |\underline{t}_{[0,m_n]}|_{\mathsf{0}} + (m_n-1)/2 \end{align*} $$

and so

$$ \begin{align*} |\underline{t}_{[0,m_{n+1}]}|_{\mathsf{1}}-|\underline{t}_{[0,m_{n+1}]}|_{\mathsf{0}} &= |\underline{t}_{[0,m_n]}|_{\mathsf{1}} -|\underline{t}_{[0,m_n]}|_{\mathsf{0}} +1 \\ & = n+2, \end{align*} $$

as we wanted. A similar argument proves the claim for $n+1$ under the assumption that the claim is true for some odd $n\in \mathbb {N}$ .

Since $\tilde {x}_{[0,\infty )}=t$ , by (3.27) and (3.25), we see that (3.24) is violated for this particular g and thus $\sigma _*$ does not act as the identity on the equivalence class of $\chi _{[\mathsf {1}]}$ , which completes the proof.

Proof. By Proposition 3.4 together with [Reference Putnam39, Theorem 1.1], we get ${C(\underline {X},\mathbb {Z})}/{\mathrm {im}(1-\varphi _*)} \cong \mathrm {K}_0(C(\underline {X})\rtimes _\varphi \mathbb {Z})\cong \mathbb {Z}[\tfrac {1}{2}]\oplus \mathbb {Z}$ with an isomorphism that carries $[1]$ to $(1,0)$ . Under this isomorphism, $\sigma _*$ becomes an order $2$ automorphism on $\mathbb {Z}[\tfrac {1}{2}]\oplus \mathbb {Z}$ that maps $(1,0)$ to itself, and by $2$ -divisibility of $\mathbb {Z}[\tfrac {1}{2}]$ , it follows that $\sigma _*(q,0)=(q,0)$ for all $q\in \mathbb {Z}[\tfrac {1}{2}]$ . Moreover, there are group homomorphisms $\alpha _1\colon \mathbb {Z}\to \mathbb {Z}[\tfrac {1}{2}]$ and $\alpha _2\colon \mathbb {Z}\to \mathbb {Z}$ such that

$$ \begin{align*} \sigma_*(0,n) = (\alpha_1(n),\alpha_2(n)),\quad n\in\mathbb{Z}, \end{align*} $$

and so

$$ \begin{align*} (0,n) &= \sigma_*^2(0,n) \\ &= \sigma_*(\alpha_1(n),\alpha_2(n)) \\ &= (\alpha_1(n+\alpha_2(n)), \alpha_2^2(n)), \quad n\in\mathbb{Z}. \end{align*} $$

We then have that $\alpha _2^2=\mathrm {id}_{\mathbb {Z}}$ , and so $\alpha _2=\mathrm {id}_{\mathbb {Z}}$ or $\alpha _2=-\mathrm {id}_{\mathbb {Z}}$ . If however $\alpha _2=\mathrm {id}_{\mathbb {Z}}$ , then $2\alpha _1(n)=\alpha _1(2n)=0$ for all $n\in \mathbb {Z}$ and so $\alpha _1=0$ (by $2$ -divisibility of $\mathbb {Z}[\tfrac {1}{2}]$ ), whence $\sigma _*=\mathrm {id}$ , which is impossible by the preceding lemma. We thus have $\alpha _2=-\mathrm {id}_{\mathbb {Z}}$ ; hence, for any $(q,n)\in \mathbb {Z}[\tfrac {1}{2}]\oplus \mathbb {Z}$ , we have

$$ \begin{align*} (1+\sigma_*)(q,n) &= (q,n) + \sigma_*(q,n ) \\ &= (q,n) + (q+\alpha_1(n), -n) \\ &= (2q+\alpha_1(n),0), \end{align*} $$

which shows that $(1+\sigma _*)(\mathbb {Z}[\tfrac {1}{2}]\oplus \mathbb {Z})\subset \mathbb {Z}[\tfrac {1}{2}]\oplus 0$ . The reverse inclusion is now also immediate since $(q,0) = (1+\sigma _*)(q/2,0)$ and, thus, we conclude that $(1+\sigma _*)(\mathbb {Z}[\tfrac {1}{2}]\oplus \mathbb {Z}) = \mathbb {Z}[\tfrac {1}{2}]\oplus 0\cong \mathbb {Z}[\tfrac {1}{2}]$ as we wanted.

Proof. Clearly, A is a unital and separable $\mathrm {C}^\ast $ -algebra, which is also nuclear as the crossed product of an action of an amenable group on a compact metric space. Moreover, A is simple, as the crossed product of a free minimal action; see [Reference Archbold and Spielberg2]. By [Reference Downarowicz and Zhang17], since $\mathbb {Z}\rtimes \mathbb {Z}_2$ has polynomial growth, the underlying action is almost finite and so A is $\mathcal {Z}$ -stable by [Reference Kerr27]. (Alternatively, note that A has finite nuclear dimension by [Reference Bönicke and Li10, Corollary C] and, thus, is $\mathcal {Z}$ -stable by [Reference Winter54].) The pair $(C(\underline {X})\subset A)$ is a $\mathrm {C}^\ast $ -diagonal since the action $\mathbb {Z}\rtimes \mathbb {Z}_2\curvearrowright \underline {X}$ is free and, thus, A also satisfies the UCT by [Reference Barlak and Li3].

Note that the action $\varphi \colon \mathbb {Z}\curvearrowright \underline {X}$ is uniquely ergodic (this follows from the fact that $\varphi \colon \mathbb {Z}\curvearrowright X_\varrho $ is uniquely ergodic and that the two Cantor minimal systems are (strongly) orbit equivalent in the sense of [Reference Giordano, Putnam and Skau25], due to Propositions 3.4 and 3.2, and thus have affinely homeomorphic simplices of invariant measures [Reference Giordano, Putnam and Skau25, Theorem 2.2]). Since the invariant measures of the dynamical system $\mathbb {Z}\rtimes \mathbb {Z}_2\curvearrowright \underline {X}$ are a subset of the invariant measures of the action of the $\mathbb {Z}$ -subgroup and since any action of an amenable group admits an invariant measure, $\mathbb {Z}\rtimes \mathbb {Z}_2\curvearrowright \underline {X}$ is uniquely ergodic and, thus, A has a unique trace, since in this setting, there is an affine homeomorphism between the space of tracial states on A and the Borel probability $\mathbb {Z}\rtimes \mathbb {Z}_2$ -invariant measures on $\underline {X}$ (cf. [Reference Li and Renault32, §4.1]).

It follows from [Reference Thomsen49, Theorem 4.42] that $\mathrm {K}_1(A)=0$ , so it remains to compute the $\mathrm {K}_0$ -group. Combining Corollary 3.7 with Remark 3.8, we have that $\mathrm {K}_0(A)\cong \mathbb {Z}_2\oplus \mathbb {Z}[\tfrac {1}{2}]$ , and $[1_A]_0$ corresponds either to the element $(0,4)$ or to $(1,4)$ . Moreover, since the map $\mathbb {Z}_2\oplus \mathbb {Z}[\tfrac {1}{2}]\ni (x,y)\mapsto (x,\tfrac {1}{4}y)\in \mathbb {Z}_2\oplus \mathbb {Z}[\tfrac {1}{2}]$ is a group automorphism that maps $(0,4)$ to $(0,1)$ and $(1,4)$ to $(1,1)$ , we obtain a group isomorphism $\mathrm {K}_0(A)\cong \mathbb {Z}_2\oplus \mathbb {Z}[\tfrac {1}{2}]$ with $[1_A]_0$ identified either with $(0,1)$ or with $(1,1)$ .

Since A is simple, the ordered group $(\mathrm {K}_0(A),\mathrm {K}_0(A)_+)$ is simple and by [Reference Thomsen49, Corollary 4.33], A has tracial topological rank zero (in the sense of Lin [Reference Rørdam and Størmer42, Definition 3.3.4]), whence by [Reference Lin33], we have that the ordered group $(\mathrm {K}_0(A),\mathrm {K}_0(A)_+)$ is weakly unperforated. We can thus apply [Reference Blackadar6, Theorem 6.8.5] to conclude that $\mathrm {K}_0(A)_+$ is equal to the set

(3.36) $$ \begin{align} \{g\in\mathrm{K}_0(A): \gamma(g)>0\text{ for all }\gamma\in\mathrm{S}(\mathrm{K}_0(A),\mathrm{K}_0(A)_+,[1_A]_0)\}\cup\{0\}, \end{align} $$

where $\mathrm {S}(\mathrm {K}_0(A),\mathrm {K}_0(A)_+,[1_A]_0)$ denotes the set of states of the scaled ordered group $(\mathrm {K}_0(A),\mathrm {K}_0(A)_+,[1_A]_0)$ . If $\gamma $ is a state on $\mathrm {K}_0(A)$ , then it induces a group homomorphism $\mathbb {Z}_2\oplus \mathbb {Z}[\tfrac {1}{2}]\to \mathbb {R}$ upon composing with the isomorphism $\mathrm {K}_0(A)\cong \mathbb {Z}_2\oplus \mathbb {Z}[\tfrac {1}{2}]$ . This map is necessarily $0$ on the $\mathbb {Z}_2$ -summand since $\mathbb {R}$ is torsion free and maps either $(1,1)$ or $(0,1)$ to $1$ (depending on which element corresponds to $[1_A]_0$ ): in either case, since the map vanishes on the $\mathbb {Z}_2$ -summand, $(0,1)$ is mapped to $1$ and so this map is necessarily the canonical embedding of the $\mathbb {Z}[\tfrac {1}{2}]$ -summand into $\mathbb {R}$ . By this observation and (3.36), we conclude that $\mathrm {K}_0(A)_+$ is carried onto $\{(x,q)\in \mathbb {Z}_2\oplus \mathbb {Z}[\tfrac {1}{2}]: q>0\}\cup \{(0,0)\}$ and the proof is complete.

Proof. Let $(\varphi ,\sigma )\colon \mathbb {Z}\rtimes \mathbb {Z}_2\curvearrowright \underline {X}$ be the action of Corollary 2.5 and set $A{:=}q C(\underline {X})\rtimes _{(\varphi ,\sigma )}(\mathbb {Z}\rtimes \mathbb {Z}_2)$ . Set $B{:=}q A\otimes \mathrm {M}_{2^\infty }$ and $D{:=}q C(\underline {X})\otimes \mathrm {D}_{2^\infty }$ . The spectrum of D is the Cartesian product of the spectra of its factors, whence D has Cantor spectrum. Moreover, $(D \subset B)$ is a $\mathrm {C}^\ast $ -diagonal [Reference Barlak and Li3, Lemma 5.1].

We first show that $B\cong \mathrm {M}_{2^\infty }$ , which we do via classification theory. Note that B is unital, simple and separable, and it is moreover nuclear as the tensor product of two nuclear $\mathrm {C}^\ast $ -algebras. Since B admits a Cartan subalgebra (namely D), it satisfies the UCT [Reference Barlak and Li3]. By its definition B is $\mathrm {M}_{2^\infty }$ -stable, which implies that B is $\mathcal {Z}$ -stable, and since both A and $\mathrm {M}_{2^\infty }$ have unique trace, so does B. As abelian groups, the $\mathrm {K}$ -theory of B can be computed via the Künneth formula [Reference Blackadar6, Theorem 23.1.3]: since the $\mathrm {K}$ -theory of $\mathrm {M}_{2^\infty }$ is torsion free and since $\mathrm {K}_1(A)=\mathrm {K}_1(\mathrm {M}_{2^\infty })=0$ (by Proposition 3.9), we have

$$ \begin{align*} \mathrm{K}_1(B) \cong (\mathrm{K}_0(A)\otimes_{\mathbb{Z}}\mathrm{K}_1(\mathrm{M}_{2^\infty}))\oplus(\mathrm{K}_1(A)\otimes_{\mathbb{Z}}\mathrm{K}_0(\mathrm{M}_{2^\infty})) = 0. \end{align*} $$

In addition, using at the fourth line below that the tensor product of abelian groups is distributive with respect to direct sums and the fact that $\mathbb {Z}_2$ has $2$ -torsion, while $\mathbb {Z}[\tfrac {1}{2}]$ is $2$ -divisible, we have group isomorphisms

(4.1) $$ \begin{align} \mathrm{K}_0(B) &\cong (\mathrm{K}_0(A) \otimes_{\mathbb{Z}}\mathrm{K}_0(\mathrm{M}_{2^\infty})) \oplus(\mathrm{K}_1(A)\otimes_{\mathbb{Z}}\mathrm{K}_1(\mathrm{M}_{2^\infty})) \end{align} $$

(4.2) $$ \begin{align} &\cong \mathrm{K}_0(A) \otimes_{\mathbb{Z}}\mathrm{K}_0(\mathrm{M}_{2^\infty}) \end{align} $$

(4.3) $$ \begin{align} &\cong \big(\mathbb{Z}_2\oplus\mathbb{Z}\big[\tfrac{1}{2}\big]\big)\otimes_{\mathbb{Z}}\mathbb{Z}\big[\tfrac{1}{2}\big] \end{align} $$

(4.4) $$ \begin{align} &\cong \mathbb{Z}\big[\tfrac{1}{2}\big]\otimes_{\mathbb{Z}}\mathbb{Z}\big[\tfrac{1}{2}\big] \end{align} $$

(4.5) $$ \begin{align} &\cong \mathbb{Z}\big[\tfrac{1}{2}\big], \end{align} $$

with (4.5) the isomorphism satisfying $q_1\otimes q_2\mapsto q_1q_2$ on elementary tensors. Note that $[1_B]_0$ is mapped to the elementary tensor $[1_A]_0\otimes [1_{\mathrm {M}_{2^\infty }}]_0$ via the isomorphisms in (4.1) and (4.2) (cf. [Reference Blackadar6, §23.1] and [Reference Schochet45]), and by Proposition 3.9, this is in turn mapped to either $(0,1)\otimes 1$ or $(1,1)\otimes 1$ via the isomorphism in (4.3). In either case, this is mapped to $1\otimes 1$ via the isomorphism in (4.4), which is then mapped to $1$ via the isomorphism in (4.5).

As for the positive cone, we argue like in the proof of Proposition 3.9. Since B is a simple (and stably finite) $\mathrm {C}^\ast $ -algebra, $(\mathrm {K}_0(B), \mathrm {K}_0(B)_+)$ is a simple ordered group and as B is the tensor product of a tracially AF algebra (in the sense of Lin [Reference Lin33]) with a UHF algebra, B is also tracially AF by [Reference Lin33, Proposition 5.9], whence $(\mathrm {K}_0(B),\mathrm {K}_0(B)_+)$ is weakly unperforated. By [Reference Blackadar6, Theorem 6.8.5], we have that $\mathrm {K}_0(B)_+$ is equal to the set

$$ \begin{align*} \{g\in \mathrm{K}_0(B): \gamma(g)>0\text{ for all }\gamma\in\mathrm{S}(\mathrm{K}_0(B),\mathrm{K}_0(B)_+,[1_B]_0)\}\cup\{0\}. \end{align*} $$

Any state on $\mathrm {K}_0(B)$ induces a group homomorphism $\gamma \colon \mathbb {Z}[\tfrac {1}{2}]\to \mathbb {R}$ by composing with the isomorphisms of (4.1)–(4.5) and this will satisfy $\gamma (1)=1$ (since $[1_B]_0$ is carried to $1$ ). There is however a unique such homomorphism by $2$ -divisibility, namely the canonical embedding of $\mathbb {Z}[\tfrac {1}{2}]$ into $\mathbb {R}$ . We conclude that $\mathrm {K}_0(B)_+$ is carried to $\mathbb {Z}[\tfrac {1}{2}]_{\ge 0}$ by the isomorphisms in (4.1)–(4.5), and thus the scaled ordered group $(\mathrm {K}_0(B),\mathrm {K}_0(B)_+,[1_B]_0)$ is isomorphic to $(\mathbb {Z}[\tfrac {1}{2}],\mathbb {Z}[\tfrac {1}{2}]_{\ge 0},1)$ . By classification (see [Reference White53, Reference Winter55]), we now have that $B\cong \mathrm {M}_{2^\infty }$ .

To show that $(D \subset B)$ is not an AF diagonal, it suffices to show that $(D \subset B)$ is not conjugate to $(\mathrm {D}_{2^\infty }\subset \mathrm {M}_{2^\infty })$ (since all AF diagonals in $\mathrm {M}_{2^\infty }$ are conjugate [Reference Power38, Theorem 5.7]; also cf. [Reference Li, Liao and Winter31, Remark 4.2]). To that end, note that we have the intermediate sub- $\mathrm {C}^\ast $ -algebra

$$ \begin{align*} D \subset (C(\underline{X})\rtimes_\varphi\mathbb{Z})\otimes \mathrm{M}_{2^\infty} \subset B \end{align*} $$

and note also that $\mathrm {K}_1((C(\underline {X})\rtimes _\varphi \mathbb {Z})\otimes \mathrm {M}_{2^\infty })\cong \mathbb {Z}$ (cf. [Reference Giordano, Putnam and Skau25]). In particular, $(C(\underline {X})\rtimes _\varphi \mathbb {Z})\otimes \mathrm {M}_{2^\infty }$ is not an AF algebra. However, if C is any intermediate sub- $\mathrm {C}^\ast $ -algebra $(\mathrm {D}_{2^\infty } \subset C \subset \mathrm {M}_{2^\infty })$ , then C is necessarily AF by the main result of [Reference Archbold and Kumjian1]. This completes the proof.

Proof. The case $n=0$ is covered by the canonical AF diagonal, since $\dim _{\mathrm {diag}}(\mathrm {D}_{2^\infty }\subset \mathrm {M}_{2^\infty })=0$ (see [Reference Li, Liao and Winter31, Theorem 4.1 and Remark 4.2]).

By (the proof of) Theorem 4.1, we have a free action $\mathbb {Z}\rtimes \mathbb {Z}_2\curvearrowright X$ of the infinite dihedral group on the Cantor space such that $D{:=}q C(X)\otimes \mathrm {D}_{2^\infty }$ is a $\mathrm {C}^\ast $ -diagonal in $A{:=}q (C(X)\rtimes (\mathbb {Z}\rtimes \mathbb {Z}_2))\otimes \mathrm {M}_{2^\infty }\cong \mathrm {M}_{2^\infty }$ .

For $1\le n<\infty $ , let $D^{(n)}{:=}q D^{\otimes n} \subset A^{\otimes n}\cong \mathrm {M}_{2^\infty }^{\otimes n}\cong \mathrm {M}_{2^\infty }$ and note that each pair $(D^{(n)} \subset \mathrm {M}_{2^\infty })$ is a Cantor spectrum diagonal [Reference Barlak and Li3, Lemma 5.1]. Upon re-arranging the tensor factors and since $(\mathrm {D}_{2^\infty }^{\otimes n} \subset \mathrm {M}_{2^\infty }^{\otimes n})\cong (\mathrm {D}_{2^\infty }\subset \mathrm {M}_{2^\infty })$ ,

$$ \begin{align*} (D^{(n)}\subset \mathrm{M}_{2^\infty}) \cong (C(X^n)\otimes \mathrm{D}_{2^\infty } \subset (C(X^n)\rtimes(\mathbb{Z}\rtimes\mathbb{Z}_2)^n)\otimes \mathrm{M}_{2^\infty}) \end{align*} $$

and so by [Reference Li, Liao and Winter31, Theorem 3.1(ii)],

$$ \begin{align*} \dim_{\mathrm{diag}}(D^{(n)}\subset \mathrm{M}_{2^\infty}) \le \dim_{\mathrm{diag}}(C(X^n) \subset C(X^n)\rtimes(\mathbb{Z}\rtimes\mathbb{Z}_2)^n). \end{align*} $$

Now, by [Reference Li, Liao and Winter31, Proposition 7.2], and since $(\mathbb {Z}\rtimes \mathbb {Z}_2)^n$ is finitely generated, virtually abelian with asymptotic dimension n (note that $\mathrm {asdim}((\mathbb {Z}\rtimes \mathbb {Z}_2)^n)\le n\cdot \mathrm {asdim}(\mathbb {Z}\rtimes \mathbb {Z}_2)=n$ , where the last equality follows from the fact that $\mathbb {Z}\rtimes \mathbb {Z}_2$ is virtually $\mathbb {Z}$ and $\mathrm {asdim}(\mathbb {Z})=1$ . In addition, since $\mathbb {Z}^n$ is a subgroup of $(\mathbb {Z}\rtimes \mathbb {Z}_2)^n$ , we have that $n=\mathrm {asdim}(\mathbb {Z}^n)\le \mathrm {asdim}((\mathbb {Z}\rtimes \mathbb {Z}_2)^n)$ ),

(4.7) $$ \begin{align} \mathrm{dim}_{\mathrm{diag}}(C(X^n)\subset C(X^n)\rtimes(\mathbb{Z}\rtimes\mathbb{Z}_2)^n)=n \end{align} $$

and, thus,

(4.7i) $$ \begin{align} \dim_{\mathrm{diag}}(D^{(n)}\subset \mathrm{M}_{2^\infty})\le n. \end{align} $$

Let $\Gamma $ be the locally finite group $\bigoplus _{\mathbb {N}}\mathbb {Z}_2$ . There is a free minimal action $\Gamma \curvearrowright Y$ , with Y a Cantor space, such that $(C(Y)\subset C(Y)\rtimes \Gamma )\cong (\mathrm {D}_{2^\infty }\subset \mathrm {M}_{2^\infty })$ , see [Reference Blackadar7, II.10.4.12(iii)]. Again upon re-arranging the tensor factors and since $(\mathrm {D}_{2^\infty }^{\otimes n} \subset \mathrm {M}_{2^\infty }^{\otimes n})\cong (\mathrm {D}_{2^\infty }\subset \mathrm {M}_{2^\infty })$ ,

(4.7a) $$ \begin{align} (D^{(n)} \subset \mathrm{M}_{2^\infty}) \cong (C(X^n\times Y)\subset C(X^n\times Y)\rtimes((\mathbb{Z}\rtimes\mathbb{Z}_2)^n\times \Gamma)). \end{align} $$

Applying (4.6) with $H= (\mathbb {Z}\rtimes \mathbb {Z}_2)^n\times \{0_\Gamma \} \le (\mathbb {Z}\rtimes \mathbb {Z}_2)^n\times \Gamma = G$ therein, together with (3.7a),

(4.7b) $$ \begin{align} &\dim_{\mathrm{diag}}(C(X^n)\otimes \mathrm{D}_{2^\infty} \subset (C(X^n)\rtimes(\mathbb{Z}\rtimes\mathbb{Z}_2)^n)\otimes\mathrm{D}_{2^\infty})\nonumber\\&\quad \le\dim_{\mathrm{diag}}(D^{(n)}\subset\mathrm{M}_{2^\infty}). \end{align} $$

Set $B{:=}q (C(X^n)\rtimes (\mathbb {Z}\rtimes \mathbb {Z}_2)^n)\otimes \mathrm {D}_{2^\infty }$ and let $\vartheta $ be a character on $\mathrm {D}_{2^\infty }$ . The map $\mathrm {id}\otimes \vartheta \colon B \to C(X^n)\rtimes (\mathbb {Z}\rtimes \mathbb {Z}_2)^n$ is a surjective $^*$ -homomorphism that carries $C(X^n)\otimes \mathrm {D}_{2^\infty }$ onto $C(X^n)$ , and so by [Reference Li, Liao and Winter31, Theorem 3.2(v)], we have

(4.8) $$ \begin{align} \dim_{\mathrm{diag}}(C(X^n) \subset C(X^n)\rtimes(\mathbb{Z}\rtimes\mathbb{Z}_2)^n)\le \dim_{\mathrm{diag}}(C(X^n)\otimes \mathrm{D}_{2^\infty} \subset B), \end{align} $$

whence by combining (4.7) with (4.8), we have

(4.7ii) $$ \begin{align} n \le \dim_{\mathrm{diag}}(D^{(n)} \subset \mathrm{M}_{2^\infty}). \end{align} $$

Putting (3.7i) and (3.7ii) together, we see that $\dim _{\mathrm {diag}}(D^{(n)}\subset \mathrm {M}_{2^\infty })=n$ as wanted.

Lastly, set $D^{(\infty )}{:=}q D^{\otimes \infty } \subset A^{\otimes \infty } \cong \mathrm {M}_{2^\infty }^{\otimes \infty } \cong \mathrm {M}_{2^\infty }$ , which is a $\mathrm {C}^\ast $ -diagonal arising as $(C(X^\infty \times Y) \subset C(X^\infty \times Y)\rtimes (\bigoplus _{\mathbb {N}}(\mathbb {Z}\rtimes \mathbb {Z}_2)\times \Gamma ))$ for the product of the coordinate-wise action $\bigoplus _{\mathbb {N}}(\mathbb {Z}\rtimes \mathbb {Z}_2)\curvearrowright X^\infty $ with $\Gamma \curvearrowright Y$ (cf. [Reference Barlak and Li3, Lemma 5.2]). For $n\in \mathbb {N}$ ,

$$ \begin{align*} (D^{(\infty)} \subset \mathrm{M}_{2^\infty})\cong (D^{\otimes n} \otimes D^{\otimes \infty} \subset A^{\otimes n} \otimes A^{\otimes \infty}) \end{align*} $$

and the latter pair can be written as a canonical crossed product pair of an action of $(\mathbb {Z}\rtimes \mathbb {Z}_2)^n\times \Gamma \times (\bigoplus _{\mathbb {N}}(\mathbb {Z}\rtimes \mathbb {Z}_2)\times \Gamma )$ . Applying (4.6) for the subgroup $(\mathbb {Z}\rtimes \mathbb {Z}_2)^n$ and arguing as in the preceding paragraph, we obtain $n\le \dim _{\mathrm {diag}}(D^{(\infty )} \subset \mathrm {M}_{2^\infty })$ . Since $n\in \mathbb {N}$ was arbitrary, this proves that $\dim _{\mathrm {diag}}(D^{(\infty )}\subset \mathrm {M}_{2^\infty })=\infty $ as desired.