2.1. Single-vertex rank $\mathsf k$ graphs

A countable small category Λ is called a rank $\mathsf k$ graph (or $\mathsf k$-graph) if there exists a functor $d:\Lambda \to \mathbb{N}^{\mathsf k}$ satisfying the following unique factorization property: For $\mu\in\Lambda, {\mathbf{n}}, {\mathbf{m}} \in \mathbb{N}^k$ with $d(\mu)={\mathbf{n}}+{\mathbf{m}}$, there exist unique $\beta\in d^{-1}({\mathbf{n}})$ and $\alpha\in d^{-1}({\mathbf{m}}) $ such that $\mu=\beta\alpha$. A functor $f:\Lambda_1 \to \Lambda_2$ is called a graph morphism if $d_2 \circ f=d_1$.

Let Λ be a $\mathsf{k}$-graph and ${\mathbf{n}}\in {\mathbb{N}}^k$. Set $\Lambda^{\mathbf{n}}:=d^{-1}({\mathbf{n}})$. For $\mu\in \Lambda$, we write $s(\mu)$ and $r(\mu)$ for the source and range of µ, respectively. Then Λ is said to be row-finite if $\vert v\Lambda^{{\mathbf{n}}}\vert \lt \infty$ for all $v \in \Lambda^0$ and ${\mathbf{n}} \in \mathbb{N}^{\mathsf k}$; and source-free if $v\Lambda^{{\mathbf{n}}} \neq \varnothing$ for all $v \in \Lambda^0$ and ${\mathbf{n}} \in \mathbb{N}^{\mathsf k}$. For more information about $\mathsf k$-graphs, refer to [Reference Katsura29]. In this paper, all $\mathsf k$-graphs are assumed to be row-finite and source-free. Actually, we focus on a special class of rank $\mathsf k$ graphs—single-vertex rank $\mathsf k$ graphs.

Single-vertex $\mathsf k$-graphs, at first sight, seem to be a very special class of $\mathsf k$-graphs. It turns out that they are a rather intriguing class to study. They have been systematically studied in the literature, e.g., [Reference Davidson15–Reference Davidson and Yang18]. There are close connections with the well-known Yang–Baxter equation [Reference Yang49].

Let $\{\epsilon_1,\ldots, \epsilon_{\mathsf{k}}\}$ be the standard basis of ${\mathbb{N}}^{\mathsf{k}}$, and Λ be a single-vertex rank $\mathsf{k}$ graph. For $1\le i\le \mathsf{k}$, write $ \Lambda^{\epsilon_i}:=\{{\mathbf{x}}^i_{\mathfrak{s}}:{\mathfrak{s}}\in[n_i]\}, $ where $n_i=|\Lambda^{\epsilon_i}|$. It follows from the factorization property of Λ that, for $1\le i \lt j\le \mathsf{k}$, there is a permutation $\theta_{ij}\in S_{n_i\times n_j}$ satisfying the following θ-commutation relations

\begin{align*} {\mathbf{x}}^i_{\mathfrak{s}} {\mathbf{x}}^j_{\mathfrak{t}} = {\mathbf{x}}^j_{{\mathfrak{t}}'} {\mathbf{x}}^i_{{\mathfrak{s}}'} \quad\text{if}\quad \theta_{ij}({\mathfrak{s}},{\mathfrak{t}}) = ({\mathfrak{s}}',{\mathfrak{t}}'). \end{align*}

To emphasize θ-commutation relations involved, this single-vertex $\mathsf k$-graph Λ is denoted as $\Lambda_\theta$ in this paper. So $\Lambda_\theta$ is the following (unital) semigroup

\begin{align*} \Lambda_\theta=\big\langle {\mathbf{x}}_{\mathfrak{s}}^i: {\mathfrak{s}}\in [n_i],\, 1\le i\le \mathsf{k};\, {\mathbf{x}}^i_{\mathfrak{s}} {\mathbf{x}}^j_{\mathfrak{t}} = {\mathbf{x}}^j_{{\mathfrak{t}}'} {\mathbf{x}}^i_{{\mathfrak{s}}'} \text{ whenever } \theta_{ij}({\mathfrak{s}},{\mathfrak{t}}) = ({\mathfrak{s}}',{\mathfrak{t}}') \big\rangle^+, \end{align*}

which is also occasionally written as

\begin{align*} \Lambda_\theta=\big\langle {\mathbf{x}}_{\mathfrak{s}}^i: {\mathfrak{s}}\in [n_i],\ 1\le i\le \mathsf{k}; \, \theta_{ij}, \, 1\le i \lt j\le \mathsf{k} \big\rangle^+. \end{align*}

One should notice that $\Lambda_\theta$ has the cancellation property due to the unique factorization property. It follows from the θ-commutation relations that every element $w\in \Lambda_\theta$ has the normal form $ w={\mathbf{x}}_{u_1}^1\cdots {\mathbf{x}}_{u_{\mathsf{k}}}^{\mathsf{k}} $ for some ${\mathbf{x}}_{u_i}^i\in\Lambda_\theta^{\epsilon_i{\mathbb{N}}}$ ( $1\le i\le \mathsf{k}$). Here we use the multi-index notation: ${\mathbf{x}}^i_{u_i}={\mathbf{x}}^i_{{\mathfrak{s}}_1}\cdots {\mathbf{x}}^i_{{\mathfrak{s}}_n}$ if $u_i={\mathfrak{s}}_1\cdots{\mathfrak{s}}_n$ with all ${\mathfrak{s}}_i$’s in $[n_i]$.

For $\mathsf{k}=2$, every permutation $\theta\in S_{n_1\times n_2}$ determines a single-vertex rank 2 graph. But for $\mathsf{k}\ge 3$, $\theta=\{\theta_{ij}:1\le i \lt j\le \mathsf{k}\}$ determines a rank $\mathsf k$ graph if and only if it satisfies a cubic condition (see, e.g., [Reference Davidson and Yang18, Reference Exel and Pardo23] for its definition). This cubic condition exactly provides interplay between $\mathsf k$-graphs and the Yang–Baxter equation [Reference Yang49].

Here are some examples of single-vertex $\mathsf k$-graphs which will be used later.

2.2. Self-similar single-vertex $\mathsf k$-graph C*-algebras

To unify the treatments of [Reference Johnson27] and [Reference Li and Yang38, Reference Murphy40], self-similar graphs and their C*-algebras naturally arise in [Reference Duwenig, Gillaspy, Norton, Reznikoff and Wright22] and are well studied there. Later, they are generalized to higher rank cases in [Reference Li and Yang34] and are further studied in [Reference Li33, Reference Li and Yang35].

Since this paper mainly focuses on single-vertex $\mathsf k$-graphs, we adapt the notions of [Reference Li33, Reference Li and Yang34] to our setting and simplify them accordingly.

Let $\Lambda_\theta$ be a single-vertex $\mathsf k$-graph. A bijection $\pi:\Lambda_\theta \to \Lambda_\theta$ is called an automorphism of $\Lambda_\theta$ if π preserves the degree map d. In general, an automorphism on $\Lambda_\theta$ is not necessarily a semigroup automorphism on $\Lambda_\theta$, as a semigroup. Denote by $\operatorname{Aut}(\Lambda_\theta)$ the automorphism group of Λ.

Let G be a (discrete countable) group. We say that G acts on $\Lambda_\theta$ if there is a group homomorphism φ from G to $\operatorname{Aut}(\Lambda_\theta)$. For $g\in G$ and $\mu\in\Lambda_\theta$, we often simply write $\varphi_g(\mu)$ as $g\cdot \mu$.

Let us record the following result [Reference Li and Yang34, Propositions 3.12 and 5.10], which will be used later without mentioning.

As in [Reference Li and Yang34], let γ be the gauge action of ${\mathbb{T}}^{\mathsf k}$ on ${\mathcal{O}}_{{\mathbb{Z}}, \Lambda_\theta}$:

\begin{equation*} \gamma_{\mathbf{t}}(s_\mu u_g s_\nu^*)={\mathbf{t}}^{d(\mu)-d(\nu)}s_\mu u_g s_\nu^* \end{equation*}

for all $\mu, \nu \in \Lambda_\theta$, $g\in {\mathbb{Z}}$, and ${\mathbf{t}}\in {\mathbb{T}}^{\mathsf{k}}$. The fixed point algebra, ${\mathcal{O}}_{{\mathbb{Z}}, \Lambda_\theta}^\gamma$, of γ is generated by the standard generators $s_\mu u_g s_\nu^*$ with $d(\mu)=d(\nu)$. We often write ${\mathcal{F}}$ to stand for ${\mathcal{O}}_{{\mathbb{Z}}, \Lambda_\theta}^\gamma$. More generally, for ${\mathbf{n}}\in {\mathbb{N}}^{\mathsf k}$, we define a mapping on ${\mathcal{O}}_{{\mathbb{Z}},\Lambda_\theta}$ by

\begin{equation*} \Phi_{\mathbf{n}}(x) = \int_{{\mathbb{T}}^{\mathsf k}} {\mathbf{t}}^{-{\mathbf{n}}}\gamma_{\mathbf{t}}(x)d{\mathbf{t}} \text{ for all }x \in {\mathcal{O}}_{{\mathbb{Z}},\Lambda_\theta}. \end{equation*}

Note that for $\mu,\nu \in \Lambda_\theta$, $g \in G$ we have

\begin{equation*} \Phi_{\mathbf{n}}(s_\mu u_g s_\nu^*) = \begin{cases} s_\mu u_g s_\nu^* & \text{if } d(\mu)-d(\nu) = {\mathbf{n}}, \\ 0 & \text{otherwise.} \end{cases} \end{equation*}

In particular, ${\mathcal{O}}_{{\mathbb{Z}}, \Lambda_\theta}^\gamma=\operatorname{Ran}\Phi_{\mathbf 0}$. Also $\Phi_{\mathbf 0}$ is a faithful conditional expectation from ${\mathcal{O}}_{{\mathbb{Z}},\Lambda_\theta}$ onto ${\mathcal{O}}_{{\mathbb{Z}}, \Lambda_\theta}^\gamma$.

We end this subsection by briefly recalling the periodicity of $(G, \Lambda_\theta)$. Let $(G,\Lambda_\theta)$ be a self-similar $\mathsf k$-graph. For $\mu,\nu \in \Lambda_\theta, g \in G$, the triple $(\mu,g,\nu)$ is called cycline if $\mu(g \cdot x)=\nu x$ for all $x \in s(\nu)\Lambda^\infty$. Clearly, every triple $(\mu, 1_G, \mu)$ $(\mu \in \Lambda)$ is cycline. Those cycline triples are said to be trivial. An infinite path $x \in \Lambda_\theta^\infty$ is said to be G-aperiodic if, for $g \in G, \mathbf p, \mathbf q \in \mathbb{N}^{\mathsf k}$ with $g \neq 1_G$ or $\mathbf p \neq \mathbf q$, we have $\sigma^{\mathbf p}(x) \neq g \cdot \sigma^{\mathbf q}(x)$; otherwise, x is called G-periodic. $(G,\Lambda_\theta)$ is said to be aperiodic if there exists a G-aperiodic path $x\in \Lambda_\theta^\infty$; and periodic otherwise.

2.3. Right LCM semigroup C*-algebras and their boundary quotient C*-algebras

Let us recall some basics about right LCM semigroups and their C*-algebras from [Reference Brownlowe, Larsen, Ramagge and Stammeier8]. Let P be a discrete left-cancellative semigroup. We say P is a right LCM semigroup if any two elements $x,y\in P$ with a right common multiple have a right least common multiple $z\in P$. Equivalently, P is right LCM if, for any $x,y\in P$, the intersection $xP\cap yP$ is either empty or equal to zP for some $z\in P$.

For a right LCM semigroup P, its C*-algebra $\mathrm{C}^*(P)$ defined in [Reference Li and Yang36] can be greatly simplified as follows: $\mathrm{C}^*(P)$ is the universal C*-algebra generated by isometries $\{v_p: p\in P\}$ and projections $\{e_{pP}: p\in P\}$ satisfying

(1)\begin{align} v_pv_q= v_{pq},\ v_p e_{qP} v_p^*=e_{pqP}, \ e_P=1, \ e_\varnothing=0,\ e_{pP}e_{qP}=e_{pP\cap qP} \end{align}

for all $p, q\in P$.

Recall that a subset $F\subseteq P$ is called a foundation set if it is finite and for each $p \in P$, there exists $q\in F$ such that $pP \cap qP\ne \varnothing$. Then the boundary quotient ${\mathcal{Q}}(P)$ of $\mathrm{C}^*(P)$ is the universal C*-algebra generated by isometries $\{v_p: p\in P\}$ and projections $\{e_{pP}: p\in P\}$ satisfying the relations in (1) and

\begin{equation*} \prod_{p\in F}(1-e_{pP})=0 \text{ for every foundation set}\ F\subseteq P. \end{equation*}

2.4. Semigroups from self-similar actions

Let $\Lambda_\theta$ be a single-vertex $\mathsf{k}$-graph. By $\Lambda^\epsilon$, we denote the set of all edges of $\Lambda_\theta$: $\Lambda^\epsilon=\bigcup_{i=1}^{\mathsf{k}}\{e\in \Lambda: d(e)=\epsilon_i\}$. Suppose that ${\mathbb{Z}}=\langle a \rangle$ acts on $\Lambda_\theta$ self-similarly. Then one can naturally associate a semigroup to the self-similar $\mathsf k$-graph $({\mathbb{Z}}, \Lambda_\theta)$ as follows:

(2)\begin{align} \mathsf{S}_{{\mathbb{N}}, \Lambda_\theta} :=\left\langle a, e: \begin{array}{ll} a e=a\cdot e a|_{e} &\text{if}\ e\in \Lambda_\theta^\epsilon\ \text{and}\ a|_e\ge 0, \\ a e (a|_{e})^{-1}=a\cdot e &\text{if}\ e\in \Lambda_\theta^\epsilon\ \text{and}\ a|_e \lt 0, \end{array} \; e\in\Lambda_\theta^\epsilon \right\rangle^+. \end{align}

This is the semigroup we focus on in this paper. Because of its importance, it deserves a name.

Here are another semigroup and a group which are closely related to the semigroup $\mathsf{S}_{{\mathbb{N}}, \Lambda_\theta}$:

\begin{align*} \mathsf{S}_{{\mathbb{Z}}, \Lambda_\theta} &:=\langle a, a^{-1}, e: a e=a\cdot e a|_e, \, e\in \Lambda_\theta^\epsilon \rangle^+,\\ \mathsf{G}_{{\mathbb{Z}}, \Lambda_\theta} &:=\langle a, e: a e=a\cdot e a|_e, \, e\in \Lambda_\theta^\epsilon \rangle. \end{align*}

3.1. Some basic properties

The two lemmas below will be used frequently. One can prove the first one by simple calculations, and the second one by applying remark 3.1. Their proofs are omitted here.

Since $\mathsf{S}_{{\mathbb{N}}, \mathsf{E}_n}$ is right LCM, from § 2.3 and the analysis above, one has the following

Proof. By proposition 3.8, the sets $\{a\}$ and $\{e_i: i\in [n]\}$ are foundation sets of $\mathsf{S}_{{\mathbb{N}}, \mathsf{E}_n}$. Then the map

\begin{equation*} {\mathcal{Q}}(\mathsf{S}_{{\mathbb{N}}, \mathsf{E}_n})\to{\mathcal{O}}_{{\mathbb{Z}}, \mathsf{E}_n},\ v_{e_\mu a^\ell} \mapsto s_{e_\mu} u_{a^\ell},\ e_{e_\mu a^\ell\mathsf{S}_{{\mathbb{N}}, \mathsf{E}_n}}\mapsto s_{e_\mu} s_{e_\mu}^* \end{equation*}

yields an isomorphism. The proof of ${\mathcal{Q}}(\mathsf S_{{\mathbb{Z}}, \mathsf E_n})\cong{\mathcal{O}}_{{\mathbb{Z}}, \mathsf{E}_n}$ is even simpler.\hfill▪

3.2. Pseudo-freeness of $({\mathbb{Z}}, \mathsf{E}_n)$

Let $({\mathbb{Z}}, \mathsf{E}_n)$ be a self-similar graph satisfying our standing assumption †.

3.3. The periodicity of $({\mathbb{Z}}, \mathsf{E}_n)$

In this subsection, we study the periodicity of self-similar graphs $({\mathbb{Z}}, \mathsf{E}_n)$ in detail. We first analyze the case when κ = 1, and then use it to study the general case.

Recall that κ is the number of orbits of ${\mathbb{Z}}$ on $\mathsf{E}_n$.

3.3.1. The case of κ = 1

Proposition 3.13. If κ = 1, then $({\mathbb{Z}}, \mathsf{E}_n)$ is periodic if and only if $n\!\mid\! m$.

Proof. Suppose that $m=n\ell$ for some $\ell \in {\mathbb{Z}}$. Let $x=e_{i_1}e_{i_2}\cdots\in \mathsf{E}_n^\infty$ be an infinite path. Repeatedly applying lemma 3.6 gives

\begin{equation*} a^{nk}\cdot x=a^{nk}\cdot e_{i_1} a^{n k}|_{e_{i_1}}\cdot (e_{i_2}\cdots)=e_{i_1} a^{n\ell k}\cdot(e_{i_2}\cdots)=\cdots=x \quad\text{for all}\quad k\in {\mathbb{Z}}. \end{equation*}

This shows that every infinite path $x\in \mathsf{E}_n^\infty$ is ${\mathbb{Z}}$-periodic in the sense of [Reference Li and Yang34]. So $({\mathbb{Z}}, \mathsf{E}_n)$ is periodic.

It remains to show that if $n\nmid m$ then $({\mathbb{Z}},\mathsf{E}_n)$ is aperiodic. To the contrary, assume that $({\mathbb{Z}}, \mathsf{E}_n)$ is periodic. It follows from [Reference Li33, Theorem 3.7] $({\mathbb{Z}}, \mathsf{E}_n)$ has a non-trivial cycline triple $(\mu, g, \nu)$. That is, $\mu g\cdot x=\nu x$ for all $x\in \mathsf{E}_n^\infty$ with $g\ne 0$ or $\mu\ne \nu$.

Case (a): $|\mu| \lt |\nu|$. Then there is a unique $\nu'\in \mathsf{E}_n^*\setminus \mathsf{E}_n^0$ such that $\nu=\mu\nu'$. Thus $g\cdot x=\nu'x$ and so $g\cdot x(0, |\nu'|)=\nu'$ for all $x\in \mathsf{E}_n^\infty$. This is impossible by noticing that n has to be greater than 1.

Case (b): $|\mu| \gt |\nu|$. Since $x\in \mathsf{E}_n^\infty$ is arbitrary, we replace x with $g^{-1}\cdot x$ and then apply case (a).

Case (c): $|\mu|=|\nu|$. Then $\mu=\nu$ and $g\cdot x=x$ for all $x\in \mathsf{E}_n^\infty$. Let $d:=\gcd(m,n) \gt 0$. Write $m=dm_0$ and $n=dn_0$. Since $n\nmid m$, we have $n_0\nmid m_0$. Write $x=e_{i_1}e_{i_2}\cdots$ with $i_j\in [n]$. Then

\begin{equation*} g\cdot (e_{i_1}e_{i_2}\cdots)=e_{i_1}e_{i_2}\cdots\implies g\cdot e_{i_1}=e_{i_1}, \ g|_{e_{i_1}}\cdot e_{i_2}=e_{i_2},\ \ldots \end{equation*}

Hence there is a sequence $\{k_i\}_{i\ge 1}\subseteq {\mathbb{Z}}$ of (non-zero) integers such that

\begin{equation*} g=a^{k_1 n}, \ a^{k_1 m}=a^{k_2 n},\ a^{k_2 m}=a^{k_3 n}, \ a^{k_3 m}=a^{k_4 n}, \ \ldots. \end{equation*}

So

\begin{equation*} k_1 m = k_2 n,\ k_2 m=k_3n, \ldots \end{equation*}

imply

\begin{equation*} k_1 m_0 = k_2 n_0,\ k_2 m_0=k_3n_0, \ldots. \end{equation*}

Thus one has that $m_0^p k_1 = n_0^p k_{p+1}$ for all $p\ge 1$. In particular $n_0^p\!\mid\! k_1$ for all $p\ge 1$ as $\gcd(m_0, n_0)=1$. But $n\!\nmid\! m$ implies $n_0 \gt 1$. So $k_1=0$ and hence g = 0. Then $(\mu, g, \nu)=(\mu, 0, \mu)$ is a trivial cycline triple. This is a contradiction.\hfill▪

Remark 3.14. Let $(G, \Lambda)$ be a self-similar $\mathsf{k}$-graph. As mentioned in [Reference Li and Yang34], it is easy to see that if Λ is periodic, then $(G, \Lambda)$ is periodic. But the converse is not true. Here is a class of counter-examples: It is well-known that $\mathsf{E}_n$ is aperiodic if n > 1. But proposition 3.13 shows that $({\mathbb{Z}}, \mathsf{E}_n)$ is periodic whenever $n\!\mid\! m$. Therefore, the periodicity of $(G, \Lambda)$ is more complicated than that of the ambient graph Λ.

We now determine all cycline triples of $({\mathbb{Z}}, \mathsf{E}_n)$ when κ = 1. Notice that all cycline triples are trivial if $({\mathbb{Z}}, \mathsf{E}_n)$ is aperiodic by theorem 2.8. Hence it suffices to consider periodic self-similar graphs $({\mathbb{Z}}, \mathsf{E}_n)$.

When n = 1, there is a unique infinite path. So the following is straightforward.

Lemma 3.15. If n = 1, then every triple $(\mu, a^\ell, \nu)$ is cycline.

Proposition 3.16. Suppose that κ = 1, n > 1, and $({\mathbb{Z}}, \mathsf{E}_n)$ is periodic. Then $(\mu, g, \nu)$ is cycline if and only if $\mu=\nu$ and $g=a^{\ell n}$ for some $\ell \in {\mathbb{Z}}$.

Proof. ‘If’ part is clear. For the ‘Only if’ part, assume that $(\mu, g, \nu)$ is cycline. Then

\begin{equation*} \mu g\cdot x= \nu x \text{ for all }x\in \mathsf{E}_n^\infty. \end{equation*}

If $|\mu|=|\nu|$, then $\mu=\nu$ and $g\cdot x=x$ for all $x\in \mathsf{E}_n^\infty$. As in the proof of proposition 3.13, one can see that $g=a^{\ell n}$ for some $\ell\in {\mathbb{Z}}$.

If $|\mu|\ne |\nu|$, WLOG, $|\nu| \gt |\mu|$. Then $\nu=\mu\nu'$ for some $\nu'\in \mathsf{E}_n^*\setminus\mathsf{E}_n^0$ and $g\cdot x= \nu' x$. This is impossible as n > 1.\hfill▪

Combining [Reference Li and Yang34, Theorem 6.6, Theorem 6.13] with corollary 3.10 yields

Theorem 3.17. ${\mathcal{Q}}(\mathsf{S}_{{\mathbb{N}}, \mathsf{E}_n})$ with κ = 1 satisfies UCT. It is simple iff $n\!\nmid\! m$. So it is a Kirchberg algebra iff $n\!\nmid\! m$.

3.3.2. The general case

For the general $\kappa\ge 1$, we begin with a relation between the periodicity of $({\mathbb{Z}}, \mathsf{E}_n)$ and that of its restrictions on orbits.

For simplification, let $\mathfrak N:=\operatorname{lcm}(n_i:1\le i\le \kappa)$.

Proposition 3.18. $({\mathbb{Z}}, \mathsf{E}_n)$ is periodic, if and only if the restriction $({\mathbb{Z}}, \mathsf{E}_{n_i})$ is periodic for each $1\le i\le \kappa$, if and only if $n_i\!\mid\! m_i$ for every $1\le i\le \kappa$.

Proof. If there is $1\le i\le \kappa$ such that $n_i\!\nmid\! m_i$, then $({\mathbb{Z}}, \mathsf{E}_{n_i})$ is aperiodic by proposition 3.16. Then clearly $({\mathbb{Z}}, \mathsf{E}_{n})$ is aperiodic.

Now let us assume that $n_i\!\mid\! m_i$ for all $1\le i\le \kappa$. Say $m_i=n_i \widetilde{m}_i$ with $0\ne \widetilde m_i\in{\mathbb{Z}}$ for $1\le i\le \kappa$. Then

\begin{equation*} a^{\mathfrak N} \, e_{s_1}^{i_1}\cdots e_{s_p}^{i_p}=e_{s_1}^{i_1}\cdots e_{s_p}^{i_p} \, a^{\mathfrak N\, \widetilde m_{i_1}\cdots\widetilde m_{i_p}}. \end{equation*}

Thus one can check that for arbitrary $x\in \mathsf{E}_n^\infty$ one has $a^{\mathfrak N} \cdot x= x$. Therefore, every infinite path x is ${\mathbb{Z}}$-periodic, and so $({\mathbb{Z}}, \mathsf{E}_n)$ is periodic.

We now determine all cycline triples. If n = 1, this is provided in lemma 3.15.

Proposition 3.19. If $({\mathbb{Z}}, \mathsf{E}_n)$ is periodic with n > 1, then $(\mu, g, \nu)$ is cycline if and only if $\mu=\nu$ and $g=a^{\ell \mathfrak N}$ for some $\ell \in {\mathbb{Z}}$.

Proof. ‘If’ part is clear. For the ‘Only if’ part, assume that $(\mu, g, \nu)$ is cycline. Then

\begin{equation*} \mu g\cdot x= \nu x \text{ for all }x\in \mathsf{E}_n^\infty. \end{equation*}

If $|\mu|=|\nu|$, then $\mu=\nu$ and $g\cdot x=x$ for all $x\in \mathsf{E}_n^\infty$. It is now not hard to see that $g=a^{\ell \mathfrak N}$ for some $\ell\in {\mathbb{Z}}$.

If $|\mu|\ne |\nu|$, WLOG, $|\nu| \gt |\mu|$. Then $\nu=\mu\nu'$ for some $\nu'\in \mathsf{E}_n^*\setminus\mathsf{E}_n^0$ and $g\cdot x= \nu' x$. This is impossible as n > 1.\hfill▪

Combining [Reference Li and Yang34, Theorem 6.6, Theorem 6.13] with corollary 3.10 yields

Theorem 3.20. ${\mathcal{O}}_{{\mathbb{Z}}, \mathsf{E}_n}$ satisfies UCT. It is simple iff $n_i\!\nmid\! m_i$ for some $1\le i\le \kappa$. So it is a Kirchberg algebra iff $n_i\!\nmid \! m_i$ for some $1\le i\le \kappa$.

Remark 3.21. It is well-known that the roles of n and m in BS groups $\operatorname{BS}(n,m)$ are symmetric in the sense of $\operatorname{BS}(n,m)\cong \operatorname{BS}(m,n)$. So $\mathrm{C}^*(\operatorname{BS}(n,m))\cong \mathrm{C}^*(\operatorname{BS}(m,n))$. However, the symmetry is lost for BS semigroups. For instance, If $0 \lt n\ne m\in {\mathbb{N}}$ satisfies $n\!\mid\!m$, then ${\mathcal{Q}}(\operatorname{BS}^+(n,m))$ is not simple while ${\mathcal{Q}}(\operatorname{BS}^+(m,n))$ is simple.

3.4. Cartan subalgebras of ${\mathcal{O}}_{{\mathbb{Z}}, \mathsf{E}_n}$

We begin with the definition of Cartan subalgebras. Let ${\mathcal{B}}$ be an abelian C*-subalgebra of a given C*-algebra ${\mathcal{A}}$. ${\mathcal{B}}$ is called a Cartan subalgebra in ${\mathcal{A}}$ if

1. (i) ${\mathcal{B}}$ contains an approximate unit in ${\mathcal{A}}$;

2. (ii) ${\mathcal{B}}$ is a MASA;

3. (iii) ${\mathcal{B}}$ is regular: the normalizer set $N({\mathcal{B}})=\{x\in {\mathcal{A}}: x{\mathcal{B}} x^*\cup x^*{\mathcal{B}} x\subseteq {\mathcal{B}}\}$ generates ${\mathcal{A}}$;

4. (iv) there is a faithful conditional expectation ${\mathcal{E}}$ from ${\mathcal{A}}$ onto ${\mathcal{B}}$.

In this subsection, we show that there is a canonical Cartan subalgebra for each ${\mathcal{O}}_{{\mathbb{Z}}, \mathsf E_n}$. It is closely related to the fixed point algebra ${\mathcal{O}}_{{\mathbb{Z}}, \mathsf{E}_n}^\gamma$ of the gauge action γ. However, there is an essential difference between the cases of n = 1 and n > 1.

3.4.2. The case of n > 1

Recall that $\mathfrak N =\operatorname{lcm}(n_i:1\le i\le \kappa)$ and $n=\sum\limits_{i=1}^\kappa n_i$.

Proposition 3.23. If n > 1, then the cycline C*-subalgebra ${\mathcal{M}}:=\mathrm{C}^*(s_\mu u_{a^{\ell\mathfrak{N}}} s_\mu^*: \mu \in \mathsf{E}_n^*, \ell\in {\mathbb{Z}})$ is a MASA in ${\mathcal{O}}_{{\mathbb{Z}}, \mathsf{E}_n}$.

Proof. We first show that ${\mathcal{M}}$ is abelian. Compute

\begin{align*} &(s_\mu u_{a^{K\mathfrak{N}}}s_\mu^*)(s_\nu u_{a^{L\mathfrak{N}}}s_\nu^*)\\ &= \begin{cases} s_\mu u_{a^{(K+L)\mathfrak{N}}}s_\mu^*& \text{if }\mu=\nu,\\ s_\mu u_{a^{K\mathfrak{N}}}s_{\nu'} u_{a^{L\mathfrak{N}}}s_\nu^*=s_\mu s_{\nu'} u_{a^{K\mathfrak{N}}|_{\nu'}}u_{a^{L\mathfrak{N}}}s_\nu^* =s_\nu u_{a^{K\mathfrak{N}}|_{\nu'}}u_{a^{L\mathfrak{N}}}s_\nu^* & \text{if }\nu=\mu\nu',\\ s_\mu u_{a^{K\mathfrak{N}}}s_{\mu'}^* u_{a^{L\mathfrak{N}}}s_\nu^*=s_\mu u_{a^{K\mathfrak{N}}} u_{(a^{-L{\mathbf{n}}}|_{a^{L\mathfrak{N}}\cdot\mu'})^{-1}}s_{\mu'}^* s_\nu^* & \text{if }\mu=\nu\mu',\\ =s_\mu u_{a^{K\mathfrak{N}}} u_{a^{L\mathfrak{N}}|_{\mu'}}s_{\mu}^*&\\ 0&\text{otherwise}. \end{cases} \end{align*}

Similar calculations yield

\begin{align*} (s_\nu u_{a^{L\mathfrak{N}}}s_\nu^*)(s_\mu u_{a^{K\mathfrak{N}}}s_\mu^*) = \begin{cases} s_\nu u_{a^{(K+L)\mathfrak{N}}}s_\nu^* &\text{if }\mu=\nu,\\ s_\nu u_{a^{L\mathfrak{N}}} u_{a^{K\mathfrak{N}}|_{\nu'}}s_{\nu}^*& \text{if }\nu=\mu\nu',\\ s_\mu u_{a^{L\mathfrak{N}}|_{\mu'}}u_{a^{K\mathfrak{N}}}s_\mu^*& \text{if }\mu=\nu\mu',\\ 0&\text{otherwise}. \end{cases} \end{align*}

Thus ${\mathcal{M}}$ is abelian.

As in [Reference Li33], let ${\mathcal{G}}_{{\mathbb{Z}}, \mathsf{E}_n}$ be the groupoid associated with the self-similar graph $({\mathbb{Z}}, \mathsf{E}_n)$. It follows from [Reference Li33, Lemma 5.2] and proposition 3.19 that $\text{Iso}({\mathcal{G}}_{{\mathbb{Z}}, \mathsf{E}_n})^\circ=\bigcup\limits_{\mu\in \mathsf{E}_n^*, \ell\in {\mathbb{Z}}} Z(\mu, a^{\ell\mathfrak{N}}, \mu)$. Also notice that ${\mathcal{M}}\cong \mathrm{C}^*(\text{Iso}({\mathcal{G}}_{{\mathbb{Z}}, \mathsf{E}_n})^\circ)$. So $\mathrm{C}^*(\text{Iso}({\mathcal{G}}_{{\mathbb{Z}}, \mathsf{E}_n})^\circ)$ is abelian. Hence, by [Reference Bruce and Scarparo10, Corollary 5.4], ${\mathcal{M}}$ is a MASA.

Remark 3.24. By lemma 3.15, when n = 1, the cycline subalgebra of ${\mathcal{O}}_{{\mathbb{Z}}, \mathsf{E}_1}$ coincides with ${\mathcal{O}}_{{\mathbb{Z}}, \mathsf{E}_1}$, which is generally not abelian. So proposition 3.23 does not hold true for n = 1.

Remark 3.25. Keep the same notation in the proof above. If $\text{Iso}({\mathcal{G}}_{{\mathbb{Z}}, \mathsf{E}_n})^\circ$ is closed, then applying [Reference Bratteli and Robinson6, Corollary 4.5] one can conclude that ${\mathcal{M}}$ is Cartan in ${\mathcal{O}}_{{\mathbb{Z}}, \mathsf{E}_n}$. But, unfortunately, $\text{Iso}({\mathcal{G}}_{{\mathbb{Z}}, \mathsf{E}_n})^\circ$ needn’t be closed in general.

3.5. Some old examples revisited

Recall the flip (single-vertex) rank 2 graphs

\begin{equation*} \Lambda_{\text{flip}}=\langle{\mathbf{e}}_i, {\mathbf{f}}_j: {\mathbf{e}}_i {\mathbf{f}}_j = {\mathbf{f}}_i {\mathbf{e}}_j, i, j\in [n]\rangle^+ \ (n\ge 2) \end{equation*}

and the square rank 2 graph

\begin{equation*} \Lambda_{\text{square}} = \langle {\mathbf{x}}_i, {\mathbf{y}}_j: {\mathbf{x}}_i {\mathbf{y}}_j={\mathbf{y}}_{i+1} {\mathbf{x}}_j, i, j\in [2] \rangle^+. \end{equation*}

Example 3.26. In [Reference Chen and Li12], it is shown that ${\mathcal{Q}}(\operatorname{BS}^+(n,n))\cong {\mathrm{C}}({\mathbb{T}}) \otimes {\mathcal{O}}_n$. From what we have obtained so far, we can prove this by relating to rank 2 graphs. In fact, we have

\begin{equation*} {\mathcal{Q}}(\operatorname{BS}^+(n,n))\cong {\mathrm{C}}({\mathbb{T}}) \otimes {\mathcal{O}}_n\cong {\mathcal{O}}_{\Lambda_{\text{flip}}}. \end{equation*}

To see this, we construct an explicit isomorphism from ${\mathcal{O}}_{\Lambda_{\text{flip}}}$ onto ${\mathcal{Q}}(\operatorname{BS}^+(n,n))$. Let $\pi: {\mathcal{O}}_{\Lambda_{\text{flip}}}\to {\mathcal{Q}}(\operatorname{BS}^+(n,n))$ be the homomorphism determined by

(3)\begin{align} s_{{\mathbf{e}}_i}\mapsto E_i:=s_{e_i}, \ s_{{\mathbf{f}}_j}\mapsto F_j:=u_{a^n} s_{e_j} \ (i, j\in [n]). \end{align}

It is easy to see that $E_iF_j=F_iE_j$ as $u_{a^n}$ is in the center of ${\mathcal{Q}}(\operatorname{BS}^+(n,n))$. Also π is surjective as from $a e_i=e_{i+1}$ $(0\le i\le n-2$) and $a e_{n-1}=e_0 a^n$ one has

\begin{align*} \sum_{i=0}^{n-1}u_a s_{e_i} s_{e_i}^* &=\sum_{i=0}^{n-2} s_{e_{i+1}} s_{e_i}^* + s_{e_0} u_{a^n} s_{e_{n-1}}^* =\sum_{i=0}^{n-2} s_{e_{i+1}} s_{e_i}^* + u_{a^n} s_{e_0} s_{e_{n-1}}^*\\ &=\sum_{i=0}^{n-2} s_{e_{i+1}} s_{e_i}^* + F_0 s_{e_{n-1}}^*. \end{align*}

Conversely, define $\rho: {\mathcal{Q}}(\operatorname{BS}^+(n,n))\to {\mathcal{O}}_{\Lambda_{\text{flip}}}$ by

\begin{equation*} s_{e_i}\mapsto s_{{\mathbf{e}}_i}, \ u_a\mapsto \sum_{i=0}^{n-2} s_{{\mathbf{e}}_{i+1}}s_{{\mathbf{e}}_i}^* + s_{{\mathbf{f}}_0} s_{{\mathbf{e}}_{n-1}}^*. \end{equation*}

Then ρ determines a homomorphism. Also one can check that π and ρ are the inverse to each other. Therefore one has $ {\mathcal{Q}}(\operatorname{BS}^+(n,n))\cong {\mathcal{O}}_{\Lambda_{\text{flip}}}$, which is also isomorphic to ${\mathrm{C}}({\mathbb{T}}) \otimes {\mathcal{O}}_n$ by [Reference Davidson, Power and Yang17].

Example 3.27. In this example, through $\operatorname{BS}^+(2,2)$, we are able to show that ${\mathcal{O}}_{\Lambda_{\text{square}}}\cong {\mathcal{O}}_{\Lambda_{\text{flip}}}$, which seems unclear in [Reference Davidson, Power and Yang17] although both ${\mathcal{O}}_{\Lambda_{\text{flip}}}$ and ${\mathcal{O}}_{\Lambda_{\text{square}}}$ are well-studied there.

Let $W:=s_{{\mathbf{e}}_1} s_{{\mathbf{e}}_0}^*+ s_{{\mathbf{f}}_0} s_{{\mathbf{e}}_1}^*$. Then $W^2=s_{{\mathbf{f}}_0} s_{{\mathbf{e}}_0}^*+s_{{\mathbf{f}}_1} s_{{\mathbf{e}}_1}^*$. Define $\pi: {\mathcal{O}}_{\Lambda_{\text{square}}}\to {\mathcal{O}}_{\Lambda_{\text{flip}}}$ via

\begin{equation*} s_{{\mathbf{x}}_0}\mapsto s_{{\mathbf{e}}_0},\ s_{{\mathbf{x}}_1}\mapsto s_{{\mathbf{e}}_1} W^*,\ s_{{\mathbf{y}}_0}\mapsto W s_{{\mathbf{e}}_0},\ s_{{\mathbf{y}}_1}\mapsto W s_{{\mathbf{e}}_1} W^*. \end{equation*}

Then one can verify that π is a homomorphism.

Let $F:=s_{{\mathbf{y}}_1} s_{{\mathbf{x}}_1}^* + s_{{\mathbf{y}}_0} s_{{\mathbf{x}}_0}^*$. Then $F^2=\sum_{i, j\in [2]} s_{{\mathbf{y}}_i{\mathbf{y}}_j} s_{{\mathbf{x}}_{(i+1)}{\mathbf{x}}_j}^*$. Let $\rho: {\mathcal{O}}_{\Lambda_{\text{flip}}}\to {\mathcal{O}}_{\Lambda_{\text{square}}}$ be defined as

\begin{equation*} s_{{\mathbf{e}}_0}\mapsto s_{{\mathbf{x}}_0},\ s_{{\mathbf{e}}_1}\mapsto s_{{\mathbf{x}}_1} F,\ s_{{\mathbf{f}}_0}\mapsto F^2 s_{{\mathbf{x}}_0}=s_{{\mathbf{x}}_0} F^2,\ s_{{\mathbf{f}}_1}\mapsto F^2 s_{{\mathbf{x}}_1} F=s_{{\mathbf{x}}_1} F^3 . \end{equation*}

Then ρ is a homomorphism. Moreover, π and ρ are the inverse to each other, and $\rho(W)=F$ and $\pi(F)=W$.

4.1. Higher rank BS semigroups

Consider $\mathsf{k}$ given self-similar graphs $({\mathbb{Z}}, \mathsf{E}_{n_i})$ with $\mathsf{E}_{n_i}=\{{\mathbf{x}}_{\mathfrak{s}}^i: s\in [n_i]\}$ ( $1\le i\le \mathsf k$). Suppose that ${\mathbf{x}}^i_{\mathfrak{s}}$’s satisfy the commutation relations $\theta_{ij}({\mathbf{x}}_{\mathfrak{s}}^i, {\mathbf{x}}_{\mathfrak{t}}^j)= ({\mathbf{x}}_{{\mathfrak{t}}'}^j , {\mathbf{x}}_{{\mathfrak{s}}'}^i)$ for $1 \leq i \lt j \leq \mathsf{k}$. Applying [Reference Laca, Raeburn, Ramagge and Whittaker32, Proposition 4.1], one has

Proposition 4.1. Keep the same notation. The $\mathsf k$ self-similar graphs $({\mathbb{Z}}, \mathsf{E}_{n_i})$ with the commutation relations θ_ij’s determine a self-similar $\mathsf k$-graph $({\mathbb{Z}}, \Lambda_\theta)$ if and only if

(4)\begin{align} (a\cdot {\mathbf{x}}_{\mathfrak{s}}^i)(a|_{{\mathbf{x}}_{\mathfrak{s}}^i} \cdot {\mathbf{x}}_{\mathfrak{t}}^j) &= (a \cdot {\mathbf{x}}_{{\mathfrak{t}}'}^j)(a|_{{\mathbf{x}}_{{\mathfrak{t}}'}^j} \cdot {\mathbf{x}}_{{\mathfrak{s}}'}^i) \end{align}

for all $1 \leq i \lt j \leq \mathsf{k}$, ${\mathfrak{s}}\in [n_i]$, ${\mathfrak{t}}\in [n_j]$.

Based on example 3.3, it is reasonable to introduce the following notion.

Here are some examples of higher rank BS semigroups.

In the sequel, we provide some interesting C*-algebras studied in the literature, which can be realized from higher rank BS semigroups.

1. (1) It is shown in [Reference Laca, Raeburn, Ramagge and Whittaker32] that the Cuntz algebra ${\mathcal{Q}}_{\mathbb{N}}$ is isomorphic to the boundary quotient of a semigroup of the form in example 4.3.

2. (2) For $2\le p\in {\mathbb{N}}$, the p-adic C*-algebra ${\mathcal{Q}}_p$ can also be recovered from a semigroup of the form in example 4.3 (cf. [Reference Laca, Raeburn, Ramagge and Whittaker32] for p = 2 in terms of standard product of odometers).

3. (3) ${\mathcal{Q}}_{\mathbb{N}}$ can be also realized as a boundary quotient of the left ax + b semigroup ${\mathbb{N}}\rtimes {\mathbb{N}}^\times$ studied in [Reference Laca and Raeburn31]. The C*-algebra of ${\mathbb{N}}\rtimes {\mathbb{N}}^\times$ itself is also related to higher rank BS semigroups.

4. (4) The boundary $\partial {\mathcal{T}}({\mathbb{N}}^\times \lt imes {\mathbb{N}})$ of the right ax + b semigroup ${\mathbb{N}}^\times \lt imes {\mathbb{N}}$ is a boundary quotient of a higher-rank BS semigroup of the form given in example 4.4.

5. (5) The C*-algebra ${\mathcal{O}}(E_{n,m})$ studied by Katsura in [Reference Katsura28] is isomorphic to ${\mathcal{O}}_{\mathsf E(n,m)}$ (see examples 3.2 and 3.3).

4.2. Relation to Furstenberg’s $\times p, \times q$ conjecture

We first give a very brief introduction on Furstenberg’s $\times p, \times q$ conjecture. For all undefined notions or any further information, refer to [Reference Brownlowe, Ramagge, Robertson and Whittaker9, Reference Fowler and Sims24, Reference Furstenberg25, Reference Nekrashevych42].

Let $2\le p,q\in {\mathbb{N}}$ be multiplicatively independent, i.e., $\frac{\ln p}{\ln q}\not\in {\mathbb{Q}}$. Define $T_p: {\mathbb{T}}\to {\mathbb{T}}$ by $T_p(z)=z^p$ for all $z\in {\mathbb{T}}$. Similarly for T_q. A subset of ${\mathbb{T}}$ is said to be $\times p, \times q$-invariant if it is invariant under both T_p and T_q. Furstenberg classifies all closed $\times p, \times q$-invariant subsets of ${\mathbb{T}}$ in [Reference Fowler and Sims24]: Such a subset is either finite or ${\mathbb{T}}$ itself. Then he conjectures the following:

Conjecture (Furstenberg’s $\times p, \times q$ conjecture). An ergodic $\times p, \times q$-invariant Borel probability measure of ${\mathbb{T}}$ is either finitely supported or the Lebesgue measure.

According to our best knowledge, this conjecture is still open. The best known result so far is the following theorem, which is proved by Rudolph when p and q are coprime in [Reference Nekrashevych41] and later improved by Johnson in [Reference Huang and Wu26].

Theorem (Rudolph–Johnson). If µ is an ergodic $\times p, \times q$-invariant measure on ${\mathbb{T}}$, then either both entropies of T_p and T_q with respect to µ are 0, or µ is the Lebesgue measure.

When both entropies of T_p and T_q with respect to µ are 0, Furstenberg’s $\times p, \times q$ conjecture is reduced to studying the C*-algebra $\mathrm{C}^*(G)$ of the group G, where

(5)\begin{align} G:=\langle s, t, z: st=ts, \ sz=z^p s,\ tz=z^q t\rangle. \end{align}

It turns out that $\mathrm{C}^*(G)\cong \mathrm{C}^*({\mathbb{Z}}[\frac{1}{pq}])\rtimes {\mathbb{Z}}^2\cong \mathrm{C}^*({\mathbb{Z}}[\frac{1}{pq}]\rtimes {\mathbb{Z}}^2)$. In [Reference Furstenberg25], the representation theory of $\mathrm{C}^*({\mathbb{Z}}[\frac{1}{pq}])\rtimes {\mathbb{Z}}^2$ is studied. In particular, the authors focus on which kind of its representations are induced by $\times p, \times q$-invariant measures on ${\mathbb{T}}$. Later, the following equivalence is shown in [Reference Brownlowe, Ramagge, Robertson and Whittaker9]: Furstenberg’s $\times p, \times q$ conjecture holds true if and only if the canonical trace is the only faithful extreme tracial state on $\mathrm{C}^*(G)\cong \mathrm{C}^*({\mathbb{Z}}[\frac{1}{pq}]\rtimes {\mathbb{Z}}^2)$.

Our purpose here is to connect Furstenberg’s conjecture with higher rank BS semigroups. For this, from (5) one has

\begin{align*} t^{-1}s^{-1}=s^{-1}t^{-1}, \ z^{-1}s^{-1}=s^{-1}z^{-p} ,\ z^{-1}t^{-1}=t^{-1}z^{-q}. \end{align*}

Let $\tilde G$ be the group

(6)\begin{align} \tilde G:=\langle \tilde s, \tilde t, \tilde z: \tilde t\tilde s=\tilde s\tilde t, \ \tilde z\tilde s=\tilde s \tilde z^{p} ,\ \tilde z \tilde t=\tilde t \tilde z^{q}\rangle. \end{align}

Thus $G\cong \tilde G$ and so $\mathrm{C}^*(G)\cong \mathrm{C}^*(\tilde G)$. The upshot by doing so is $\mathrm{C}^*(G)\cong {\mathcal{O}}_{\Lambda_{\mathsf d}((1,1),(p,q))}$.

Now return to (5) again. Let $G^+$ be the corresponding semigroup

(7)\begin{align} G^+:=\langle s, t, z: st=ts, \ sz=z^p s,\ tz=z^q t\rangle^+. \end{align}

Then we claim that ${\mathcal{Q}}(G^+)\cong {\mathcal{O}}_{\Lambda_{\mathsf d}((p,q),(1,1))}$. In fact, let $e_i:=z^i s$ and $f_j:=z^j t$ for $i\in [p]$ and $j\in [q]$. Then $st=ts\iff e_0 f_0=f_0e_0$. For $k\in [p]$ and $\ell \in [q]$, let $k'\in [p]$ and $\ell'\in [q]$ be the unique ones such that $k+\ell p= \ell' + k' q$. Then we have

\begin{align*} e_k f_\ell =z^k s z^\ell t=z^k z^{\ell p} st= z^{\ell'} z^{k' q} ts=z^{\ell'} tz^{k'}s=f_{\ell'} e_{k'}. \end{align*}

Thus there is a homomorphism

\begin{equation*} \pi: {\mathcal{O}}_{\Lambda_{\mathsf d}((p,q),(1,1))} = {\mathcal{O}}_{{\mathbb{Z}}, \Lambda_{\mathsf d}} \to {\mathcal{Q}}(G^+), \ s_{{\mathbf{x}}_i^1}\mapsto v_{e_i},\ s_{{\mathbf{x}}_j^2}\mapsto v_{f_j},\ u_a\mapsto v_z, \end{equation*}

which is an isomorphism as it has an inverse given by

\begin{equation*} v_s\mapsto s_{{\mathbf{x}}^1_0},\ v_t\mapsto s_{{\mathbf{x}}^2_0},\ v_z\mapsto u_a. \end{equation*}

Hence ${\mathcal{O}}_{\Lambda_{\mathsf d}((p,q),(1,1))} \cong {\mathcal{Q}}(G^+)$.

To sum up, we have shown that

(8)\begin{align} \mathrm{C}^*(G)\cong {\mathcal{O}}_{\Lambda_{\mathsf d}((1,1),(p,q))} \text{ and }{\mathcal{Q}}(G^+)\cong {\mathcal{O}}_{\Lambda_{\mathsf d}((p,q),(1,1))}. \end{align}

Based on the above, in what follows, we focus on two extreme, but rather interesting, classes of higher rank BS semigroups: $\Lambda_{\mathsf d}(\mathfrak n, \mathbb{ 1}_{\mathsf k})$ with $\mathbb{ 1}_{\mathsf k}\le \mathfrak n\in {\mathbb{N}}^{\mathsf k}$ and $\Lambda_{\mathsf d}(\mathbb{ 1}_{\mathsf k}, \mathfrak m)$ with $m_i\ne 0$ for all $1\le i\le \mathsf k$.

From now on, to unify our notation, the set $\{0,1\ne p_i \in {\mathbb{Z}}: 1\le i\le \mathsf k\}$ is said to be multiplicatively independent if there is no $\mathbf 0\ne q\in {\mathbb{Z}}^{\mathsf k}$ such that $\prod\limits_{i=1}^{\mathsf k} p_i^{q_i}=1$. When all p_i’s are also $\ge 1$, $\{p_i: 1\le i\le \mathsf k\}$ is multiplicatively independent if and only if $\{\ln p_i: 1\le i\le \mathsf k\}$ is rationally independent.

4.3. The case of $\mathfrak m=\mathbb{ 1}_{\mathsf k}$

The C*-algebra of the self-similar $\mathsf k$-graph $\Lambda_{\mathsf d} (\mathfrak n, \mathbb{ 1}_{\mathsf k})$ is studied in [Reference Laca, Raeburn, Ramagge and Whittaker32]. It is shown there that ${\mathcal{O}}_{\Lambda_{\mathsf d}(\mathfrak n, \mathbb{ 1}_{\mathsf k})}$ is simple if and only if $\{n_i: 1\le i\le \mathsf k\}$ is multiplicatively independent, and that its cycline subalgebra is Cartan by [Reference Li33, Theorem 5.6]. In what follows, we identify the center of ${\mathcal{O}}_{\Lambda_{\mathsf d}(\mathfrak n, \mathbb{ 1}_{\mathsf k})}$, which is overlooked in [Reference Laca, Raeburn, Ramagge and Whittaker32, Reference Li33]. A useful lemma first:

By [Reference Li33, Theorem 7.5] (or Appendix there), $\Lambda_{\mathsf d}(\mathfrak n, \mathbb{ 1}_{\mathsf k})$ and $\Lambda_{\mathsf d}$ share the same periodicity $\{\mathbf p\in {\mathbb{Z}}^{\mathsf k}: \mathfrak n^{\mathbf p}=1\}$. Thus, for each pair $(\mathbf p, \mathbf q)\in {\mathbb{N}}^{\mathsf k}\times {\mathbb{N}}^{\mathsf k}$ with $\mathfrak n^{\mathbf p}=\mathfrak n^{\mathbf q}$, there is a bijection $\phi_{\mathbf p, \mathbf q}: \Lambda_{\mathsf d}^{\mathbf p} \to \Lambda_{\mathsf d}^{\mathbf q} $ satisfying

(9)\begin{align} \mu \nu = \phi_{\mathbf p, \mathbf q}(\mu) \phi_{\mathbf p, \mathbf q}^{-1}(\nu),\ \phi_{\mathbf p, \mathbf q}^{-1}(\nu)\phi_{\mathbf p, \mathbf q}(\mu)=\nu \mu \end{align}

for every pair $(\mu, \nu)\in \Lambda_{\mathsf d}^{\mathbf p}\times \Lambda_{\mathsf d}^{\mathbf q}$ [Reference Davidson and Yang18, Theorem 7.1] (or [Section 5]). Let

\begin{equation*} V_{\mathbf p, \mathbf q}:=\sum_{\mu\in \Lambda_{\mathsf d}^{\mathbf p}} s_\mu s_{\phi_{\mathbf p, \mathbf q}(\mu)}^*. \end{equation*}

By [Reference Davidson and Yang18, Theorem 4.9], each $V_{\mathbf p, \mathbf q}$ is a unitary in ${\mathcal{O}}_{\Lambda_{\mathsf d}(\mathfrak n, \mathbb{ 1}_{\mathsf k})}$.

Proposition 4.8. The center of ${\mathcal{O}}_{\Lambda_{\mathsf d}(\mathfrak n, \mathbb{ 1}_{\mathsf k})}$ is given by

\begin{equation*} {\mathcal{Z}}({\mathcal{O}}_{\Lambda_{\mathsf d}(\mathfrak n, \mathbb{ 1}_{\mathsf k})}) =\mathrm{C}^*(V_{\mathbf p, \mathbf q}: (\mathbf p, \mathbf q)\in {\mathbb{N}}^{\mathsf k}\times {\mathbb{N}}^{\mathsf k},\ \mathfrak n^{\mathbf p}=\mathfrak n^{\mathbf q}). \end{equation*}

In particular, ${\mathcal{Z}}({\mathcal{O}}_{\Lambda_{\mathsf d}(\mathfrak n, \mathbb{ 1}_{\mathsf k})})$ is trivial, if and only if $\{n_i: 1\le i\le \mathsf k\}$ is multiplicatively independent.

Proof. Suppose that $A\in {\mathcal{Z}}({\mathcal{O}}_{\Lambda_{\mathsf d}(\mathfrak n, \mathbb{ 1}_{\mathsf k})})$. Then by proposition 2.7, one has $A\in {\mathcal{Z}}({\mathcal{O}}_{\Lambda_{\mathsf d}})$. But ${\mathcal{Z}}({\mathcal{O}}_{\Lambda_{\mathsf d}})=\mathrm{C}^*(V_{\mathbf p, \mathbf q}: (\mathbf p, \mathbf q)\in {\mathbb{N}}^{\mathsf k}\times {\mathbb{N}}^{\mathsf k},\ \mathfrak n^{\mathbf p}=\mathfrak n^{\mathbf q})$ [Reference Davidson and Yang18].

It remains to show that each $V_{\mathbf p, \mathbf q}$ is indeed in ${\mathcal{Z}}({\mathcal{O}}_{\Lambda_{\mathsf d}(\mathfrak n, \mathbb{ 1}_{\mathsf k})})$. This has been proved in [Reference Davidson and Yang18, Theorem 4.9]. In what follows, we prove this by invoking lemma 4.7.

We first show that u_a commutes with $V_{\mathbf p, \mathbf q}$. Then on one hand, we have

\begin{align*} a\cdot (\mu \nu)=a\cdot(\phi_{\mathbf p, \mathbf q}(\mu) \phi_{\mathbf p, \mathbf q}^{-1}(\nu)) \implies a\cdot \mu a|_\mu\cdot \nu = a\cdot \phi_{\mathbf p, \mathbf q}(\mu) a|_{\phi(\mu)}\cdot \phi_{\mathbf p, \mathbf q}^{-1}(\nu). \end{align*}

On the other hand, we have

\begin{equation*} a\cdot \mu a|_\mu\cdot \nu = \phi_{\mathbf p, \mathbf q}(a\cdot \mu) \phi_{\mathbf p, \mathbf q}^{-1}(a|_\mu\cdot \nu). \end{equation*}

Thus

(10)\begin{align} & a^{-1}\cdot \phi_{\mathbf p, \mathbf q}(\mu) a^{-1}|_{\phi_{\mathbf p, \mathbf q}(\mu)}\cdot \phi_{\mathbf p, \mathbf q}^{-1}(\nu) =\phi_{\mathbf p, \mathbf q}(a^{-1}\cdot \mu) \phi_{\mathbf p, \mathbf q}^{-1}(a^{-1}|_\mu\cdot \nu) \nonumber\\ &\implies a^{-1}\cdot \phi_{\mathbf p, \mathbf q}(\mu) =\phi_{\mathbf p, \mathbf q}(a^{-1}\cdot \mu) \quad \text{and} \quad a^{-1}|_{\phi_{\mathbf p, \mathbf q}(\mu)}\cdot \phi_{\mathbf p, \mathbf q}^{-1}(\nu)=\phi_{\mathbf p, \mathbf q}^{-1}(a^{-1}|_\mu\cdot \nu) \end{align}

Then one can easily see $g\cdot \phi_{\mathbf p, \mathbf q}(\mu) = \phi_{\mathbf p, \mathbf q}(g\cdot \mu)$ for any $g\in {\mathbb{Z}}$. Similarly, one gets from the second identity of (9) that $g\cdot \phi_{\mathbf p, \mathbf q}^{-1}(\nu) = \phi_{\mathbf p, \mathbf q}^{-1}(g\cdot \nu)$ for any $g\in {\mathbb{Z}}$. Thus the second identity of (10) yields $a^{-1}|_{\phi_{\mathbf p, \mathbf q}(\mu)}=a^{-1}|_\mu$, and so $ (a^{-1}|_{\phi_{\mathbf p, \mathbf q}(\mu)})^{-1}=(a^{-1}|_\mu)^{-1}. $ Therefore

(11)\begin{align} a|_{a^{-1}\cdot \phi_{\mathbf p, \mathbf q}(\mu)}=a|_{a^{-1}\cdot \mu}. \end{align}

Now compute

\begin{align*} u_a V_{\mathbf p, \mathbf q}& = \sum_{d(\mu)=\mathbf p} s_{a\cdot \mu}u_{a|_\mu} s_{\phi_{\mathbf p, \mathbf q}(\mu)}^* = \sum_{d(\mu)=\mathbf p} s_\mu u_{a|_{a^{-1}\cdot \mu}} s_{\phi_{\mathbf p, \mathbf q}(a^{-1}\cdot\mu)}^*\\ &= \sum_{d(\mu)=\mathbf p} s_\mu u_{a|_{a^{-1}\cdot \phi_{\mathbf p, \mathbf q}(\mu)}} s_{a^{-1}\cdot\phi_{\mathbf p, \mathbf q}(\mu)}^* \ (\text{from } (10)\text{and }(11))\\ &= \sum_{d(\mu)=\mathbf p} s_\mu u_{({a^{-1}|_ {\phi_{\mathbf p, \mathbf q}(\mu)})^{-1}}} s_{a^{-1}\cdot\phi_{\mathbf p, \mathbf q}(\mu)}^* = \left(\sum_{d(\mu)=\mathbf p} s_\mu s_{\phi_{\mathbf p, \mathbf q}(\mu)}^*\right)u_a\\ &= V_{\mathbf p, \mathbf q}\, u_a . \end{align*}

Then

\begin{align*} (u_{a^m} s_\alpha s_\beta^*) V_{\mathbf p, \mathbf q} = u_{a^m} V_{\mathbf p, \mathbf q} s_\alpha s_\beta^* = V_{\mathbf p, \mathbf q} (u_{a^m} s_\alpha s_\beta^*). \end{align*}

By lemma 4.7 (ii), $\sum\limits_{d(\mu)=\mathbf p} s_\mu s_{\phi_{\mathbf p, \mathbf q}(\mu)}^*\in {\mathcal{Z}}({\mathcal{O}}_{\Lambda_{\mathsf d}(\mathfrak n, \mathbb{ 1}_{\mathsf k})})$.\hfill▪

For ${\mathbf{m}}\in {\mathbb{N}}^{\mathsf{k}}$, let ${\mathcal{F}}_{{\mathbf{m}}}:=\overline{\operatorname{span}}\{s_\mu a^n s_\nu^*: \mu,\nu\in \Lambda_{(\mathfrak n, \mathbb{ 1}_{\mathsf k})} \text{with } d(\mu)=d(\nu)={\mathbf{m}}, n\in {\mathbb{Z}}\}$, and so ${\mathcal{F}}=\overline{\bigcup\limits_{m\in {\mathbb{N}}^{\mathsf{k}}} {\mathcal{F}}_{{\mathbf{m}}}}^{\|\cdot\|}$. Also notice that, due to the Cuntz–Krieger relations, we have ${\mathcal{F}}=\lim\limits_{\stackrel{\longrightarrow}{m\in {\mathbb{N}}}}{\mathcal{F}}_{m\mathbb{1}_{\mathsf k}}$.

Let $d:=\prod\limits_{i=1}^{\mathsf{k}}\prod\limits_{p\in {\mathcal{P}}, \, p|n_i} p$, the product of all primes dividing some n_i’s.

Proof. The proof below is motivated by the proof of [Reference Li and Yang35, Theorem 3.16].

For ${\mathbf{m}} \in {\mathbb{N}}^{\mathsf k}$, let $\{e_{\mu,\nu}\}_{\mu,\nu\in \Lambda^{{\mathbf{m}}}_{\mathsf d}}$ be the matrix entries of $K(\ell^2(\Lambda_{\mathsf d}^{\mathbf{m}})).$ To simplify our notation, let $\{a^n\}_{n\in {\mathbb{Z}}}$ be the generating unitaries of $\mathrm{C}^*({\mathbb{Z}})$. Clearly, there is a homomorphism

\begin{equation*} \rho_1: K(\ell^2(\Lambda_{\mathsf d}^{m\mathbb{1}_{\mathsf k}}))\to {\mathcal{F}}_{m\mathbb{1}_{\mathsf k}},\ e_{\mu,\nu}\mapsto s_\mu s_\nu^*, \end{equation*}

and

\begin{equation*} \rho_2:{\mathrm{C}}({\mathbb{T}})\to {\mathcal{F}}_{m\mathbb{1}_{\mathsf k}},\ a^n\mapsto \sum_{\mu\in\Lambda_{\mathsf d}^{m\mathbb{1}_{\mathsf k}}} s_\mu u_{a^n} s_\mu^*. \end{equation*}

Some simple calculations show the images of ρ ₁ and ρ ₂ commute. By [Reference Li and Yang37, Theorem 6.3.7], there is a homomorphism $\rho: K(\ell^2(\Lambda_{\mathsf d}^{m\mathbb{1}_{\mathsf k}}))\otimes \mathrm{C}^*({\mathbb{Z}})\to {\mathcal{F}}_{m\mathbb{1}_{\mathsf k}}$ satisfying $\rho(e_{\mu, \nu}\otimes a^n)= s_\mu u_{a^n} s_\nu^*$. It is not hard to see that ρ is also invertible and so an isomorphism.

Set $\mu_0:={\mathbf{x}}_0^1\cdots {\mathbf{x}}^{\mathsf k}_0$ and $\mu_{\mathfrak n -\mathbb{1}}:={\mathbf{x}}^1_{n_1-1}\cdots {\mathbf{x}}^{\mathsf k}_{n_{\mathsf k}-1}$. Embed $K(\ell^2(\Lambda_{\mathsf d}^{m\mathbb{ 1}_{\mathsf k}}))\otimes \mathrm{C}^*({\mathbb{Z}})$ into $K(\ell^2(\Lambda_{\mathsf d}^{(m+1)\mathbb{ 1}_{\mathsf k}}))\otimes \mathrm{C}^*({\mathbb{Z}})$ as follows:

\begin{equation*} e_{\mu, \nu}\otimes a^n\mapsto \sum\limits_{\alpha\in \Lambda_{\mathsf d}^{\mathbb{1}}}e_{\mu a^n\cdot \alpha, \nu\alpha}\otimes a^n|_\alpha. \end{equation*}

Notice that $a|_\alpha=a$ if $\alpha=\mu_{\mathfrak n -\mathbb{1}_{\mathsf k}}$, and $a|_\alpha=0$, otherwise.

Now we have

\begin{align*} u_a &=u_a \sum_{\mu\in\Lambda^{m\mathbb{1}_{\mathsf k}}} s_\mu s_\mu^* =\sum_{\mu\in\Lambda_{\mathsf d}^{m\mathbb{1}_{\mathsf k}}} s_{a\cdot\mu} u_{a|_\mu} s_\mu^*\\ &=\sum_{\mu\ne \mu_{\mathfrak n-\mathbb{1}_{\mathsf k}}} s_{a\cdot\mu} s_\mu^*+s_{a\cdot\mu_{\mathfrak n-\mathbb{1}_{\mathsf k}}} u_a s_{\mu_{\mathfrak n-\mathbb{1}_{\mathsf k}}}^*\\ &=\sum_{\mu\ne \mu_{\mathfrak n-\mathbb{1}_{\mathsf k}}} s_{a\cdot\mu} s_\mu^*+s_{\mu_0} u_a s_{\mu_{\mathfrak n-\mathbb{1}_{\mathsf k}}}^*. \end{align*}

Therefore ${\mathcal{F}}$ is isomorphic to a Bunce–Deddens algebra of type of $d^\infty$, and so it has a unique faithful tracial state [Reference Connes and Marcolli14].\hfill▪

By lemma 4.9 and [Reference Connes and Marcolli14], ${\mathcal{F}}$ has a unique faithful tracial state, say τ, given by

\begin{equation*} \tau(s_\mu u_{a^n} s_\mu^*)=\mathfrak n^{-d(\mu)}\delta_{n,0}. \end{equation*}

Recall that $\Phi_{\mathbf 0}$ is the faithful conditional expectation from ${\mathcal{O}}_{\Lambda_{\mathsf d}(\mathfrak n, \mathbb{ 1})}$ onto ${\mathcal{F}}$ via the gauge action γ (see § 2.2). Then $\omega:=\tau\circ \Phi_{\mathbf 0}$ is a state of ${\mathcal{O}}_{\Lambda_{\mathsf d}(\mathfrak n, \mathbb{ 1}_{\mathsf k})}$.

We recall the notion of KMS states from [Reference Bartholdi, Grigorchuk and Nekrashevych5] (also see [Reference Clark, an Huef and Raeburn13]) and give some basic properties of KMS states for ${\mathcal{O}}_{\Lambda_{\mathsf d}(\mathfrak n, \mathbb{ 1}_{\mathsf k})}$.

Recall the gauge action $\gamma:{\mathbb{T}}^{\mathsf k} \to \operatorname{Aut}({\mathcal{O}}_{\Lambda_{\mathsf d}(\mathfrak n, \mathbb{ 1}_{\mathsf k})})$:

\begin{equation*} \gamma_z(s_\mu)=z^{d(\mu)}s_\mu\text{and } \gamma_z(u_g)=u_g\quad\text{for all}\quad z \in {\mathbb{T}}^{\mathsf k}, \mu \in \Lambda,g \in {\mathbb{Z}}. \end{equation*}

Let $\mathsf{r}=(\ln n_1, \ldots, \ln n_{\mathsf k})\in {\mathbb{R}}^{\mathsf k}$. Define a strongly continuous homomorphism $\alpha^{\mathsf{r}}:\mathbb{R} \to \operatorname{Aut}({\mathcal{O}}_{\Lambda_\mathsf d(\mathfrak n,\mathbb{ 1}_{\mathsf k})})$ by $\alpha^{\mathsf{r}}_t:=\gamma_{e^{it\mathsf{r}}}$. Notice that, for $\mu,\nu \in \Lambda_\mathsf d(\mathfrak n, \mathbb{ 1}_{\mathsf k}), g \in {\mathbb{Z}}$, the function $\mathbb{C} \to {\mathcal{O}}_{\Lambda_\mathsf d(\mathfrak n, \mathbb{ 1})}$, $\xi \mapsto e^{i \xi \mathsf{r} \cdot (d(\mu)-d(\nu))}s_\mu u_g s_\nu^*$ is an entire function. So $s_\mu u_g s_\nu^*$ is an analytic element. By proposition 2.7, in order to check the KMS $_\beta$ condition, it is sufficient to check whether it is valid on the set $\{s_\mu u_g s_\nu^*:\mu,\nu \in \Lambda_\mathsf d(\mathfrak n, \mathbb{ 1}_{\mathsf k}), g\in {\mathbb{Z}}\}$. In this section, we study basic properties of KMS $_\beta$ states of the one-parameter dynamical system $(\mathcal{O}_{\Lambda_\mathsf d(\mathfrak n, \mathbb{ 1})},\mathbb{R},\alpha^{\mathsf{r}})$.

Proof. An easy calculation shows

\begin{align*} \omega(s_\mu u_{a^n} s_\nu^*)=\delta_{\mu, \nu} \delta_{n,0} \, \mathfrak n^{-d(\mu)}. \end{align*}

By [Reference Li33, Theorems 6.11 and 6.12], ω is the unique KMS₁ state of ${\mathcal{O}}_{\Lambda_\mathsf d(\mathfrak n, \mathbb{ 1})}$ (also see [Reference Li33, Theorem 7.1]).\hfill▪

As in [Reference Serre46, Reference Spielberg47], let $L^2({\mathcal{O}}_{\Lambda_\mathsf d(\mathfrak n, \mathbb{ 1}_{\mathsf k})})$ be the GNS Hilbert space determined by the state ω: $\langle A, B\rangle:=\omega(A^*B)$ for all $A, B\in {\mathcal{O}}_{\Lambda_\mathsf d(\mathfrak n, \mathbb{ 1}_{\mathsf k})}$. For $A \in {\mathcal{O}}_{\Lambda_\mathsf d(\mathfrak n, \mathbb{ 1}_{\mathsf k})}$, we denote the left action of A by $\pi (A): \pi(A)B = AB$ for all $B \in {\mathcal{O}}_{\Lambda_\mathsf d(\mathfrak n, \mathbb{ 1})}$. Let $ {\mathcal{O}}_{\Lambda_\mathsf d(\mathfrak n, \mathbb{ 1}_{\mathsf k})}^c$ stand for the algebra as the finite linear span of the generators $s_\mu u_g s_\nu^*$.

Define

\begin{align*} S(A):=A^*,\ F(s_\mu u_{a^n} s_\nu^*):=\mathfrak n^{d(\mu)-d(\nu)} s_\nu u_{a^{-n}} s_\mu^*, \quad\text{for}\quad A\in {\mathcal{O}}_{\Lambda_\mathsf d(\mathfrak n, \mathbb{ 1}_{\mathsf k})}^c. \end{align*}

Then $F=S^*$. Also, if

\begin{equation*} J(s_\mu u_{a^n} s_\nu^*):=\mathfrak n^{\frac{d(\mu)-d(\nu)}{2}} s_\nu u_{a^{-n}} s_\mu^*, \ \Delta (s_\mu u_{a^n} s_\nu^*):=\mathfrak n^{d(\nu)-d(\mu)} s_\mu u_{a^n} s_\nu^*, \end{equation*}

one has

\begin{equation*} S=J\Delta^{\frac{1}{2}}=\Delta^{-\frac{1}{2}} J,\ F=J\Delta^{-\frac{1}{2}}= \Delta^{\frac{1}{2}}J. \end{equation*}

Let $\pi_\omega({\mathcal{O}}_{\Lambda_d(\mathfrak n, \mathbb{ 1}_{\mathsf k})})^{\prime\prime}$ be the von Neumann algebra generated by the GNS representation of ω. Then $\pi_\omega({\mathcal{O}}_{\Lambda_d(\mathfrak n, \mathbb{ 1}_{\mathsf k})})^{\prime\prime}$ coincides with the left von Neumann algebra of $ {\mathcal{O}}_{\Lambda_\mathsf d(\mathfrak n, \mathbb{ 1}_{\mathsf k})}^c$.

The proof of the following theorem can now be easily adapted from [Reference Serre46, Reference Spielberg47] combined with [Reference Laca, Raeburn, Ramagge and Whittaker32, Reference Li33] and is left to the interested reader.

4.4. The case of $\mathfrak n=\mathbb{ 1}_{\mathsf k}$

Since there is only one edge for each colour i, in order to ease our notation, we write ${\mathbf{x}}_i$ (instead of the notation ${\mathbf{x}}_1^i$ used above) for this unique edge. Thus ${\mathbf{x}}_i {\mathbf{x}}_j={\mathbf{x}}_j {\mathbf{x}}_i$ for all $1\le i\ne j\le \mathsf k$. Since this is the unique commutation relation on ${\mathbf{x}}_i$’s, we denote $\Lambda_\mathsf d(\mathbb{ 1}_{\mathsf k}, \mathfrak m)$ simply as $\Lambda(\mathbb{ 1}_{\mathsf k}, \mathfrak m)$. The ambient $\mathsf k$-graph is just denoted as $\Lambda_{\mathbb{ 1}_{\mathsf k}}$.

We should mention that it seems that the case of $\mathfrak n=\mathbb{ 1}_{\mathsf k}$ is also studied in [Reference Li33]. However, this is exactly the case which is completely ignored there. This could be due to two reasons: one is that $\Lambda_\mathsf d(\mathbb{ 1}_{\mathsf k}, \mathfrak m)$ is clearly not locally faithful, which is the crucial property required in [Reference Li33]; the other is that this case, at first glance, seems too special.

Our next goal is to show that $\Lambda_{\mathsf d}(\mathbb{ 1}_{\mathsf k}, \mathfrak m)$ has a canonical Cartan subalgebra. In particular, this canonical Cartan subalgebra is ${\mathcal{F}}$ if $\{|m_i|: 1\le i\le \mathsf k\}$ is multiplicatively independent; it properly contains ${\mathcal{F}}$, otherwise.

Proof. It is known and easy to see that ${\mathcal{F}}=\overline{\operatorname{span}}\{s_\mu u_{a^m} s_\nu^*\mid \mu,\nu\in \Lambda_{\mathbb{ 1}_{\mathsf k}}, d(\mu)=d(\nu), m\in {\mathbb{Z}}\}$. It follows from observation 4.14 that $d(\mu)=d(\nu)$ forces $\mu=\nu$, say equal to ${\mathbf{x}}^{\mathbf n}$ for some $\mathbf n\in {\mathbb{N}}^{\mathsf k}$. WLOG, we assume that $n_1-n_2=l \ge 0$. Then

\begin{align*} s_{{\mathbf{x}}^{\mathbf n}} u_{a^m} s_{{\mathbf{x}}^{\mathbf n}}^{-1} =&s_{{\mathbf{x}}_3^{n_3}\cdots {\mathbf{x}}_{\mathsf k}^{n_{\mathsf k}}} s_{{\mathbf{x}}_1^{n_1}} s_{{\mathbf{x}}_2^{n_2}} u_{a^m} s_{{\mathbf{x}}_2^{n_2}}^{-1} s_{{\mathbf{x}}_1^{n_1}} ^{-1} s_{{\mathbf{x}}_3^{n_3}\cdots {\mathbf{x}}_{\mathsf k}^{n_{\mathsf k}}}^{-1}\\ =&s_{{\mathbf{x}}_3^{n_3}\cdots {\mathbf{x}}_{\mathsf k}^{n_{\mathsf k}}} s_{{\mathbf{x}}_1^{n_1}} s_{{\mathbf{x}}_2^{n_2}} s_{{\mathbf{x}}_2^\ell} u_{a^m|_{{\mathbf{x}}_2^\ell}} s^{-1}_{{\mathbf{x}}_2^\ell} s_{{\mathbf{x}}_3^{n_3}\cdots {\mathbf{x}}_{\mathsf k}^{n_{\mathsf k}}}^{-1} s_{{\mathbf{x}}_2^{n_2}}^{-1} s_{{\mathbf{x}}_1^{n_1}}^{-1}\\ =&s_{{\mathbf{x}}_3^{n_3}\cdots {\mathbf{x}}_{\mathsf k}^{n_{\mathsf k}}} s_{{\mathbf{x}}_1^{n_1}} s_{{\mathbf{x}}_2^{n_1}} u_{a^m|_{{\mathbf{x}}_2^\ell}} s_{{\mathbf{x}}_3^{n_3}\cdots {\mathbf{x}}_{\mathsf k}^{n_{\mathsf k}}}^{-1} s_{{\mathbf{x}}_2^{n_1}}^{-1} s_{{\mathbf{x}}_1^{n_1}}^{-1}. \end{align*}

After repeating this process, all the ${\mathbf{x}}_i$’s will have the same exponent.\hfill▪

Proof. Compute

\begin{align*} (s_\mu u_{a^l} s_\mu^*)(s_\nu u_{a^n} s_\nu^*) &=s_\mu u_{a^l} s_\nu s_\mu^* u_{a^n} s_\nu^*\ (\text{as}\ s_\mu^*s_\nu= s_\nu s_\mu^*)\\ &=s_\mu s_\nu u_{a^l|_{\nu}} (u_{a^{-n}} s_\mu)^* s_\nu^*\ (\text{as}\ a^l\cdot \nu =\nu)\\ &=s_\mu s_\nu u_{a^l|_{\nu}} (u_{a^{-n}}|_{\mu})^{-1} s_\mu^* s_\nu^*\\ &=s_\mu s_\nu u_{a^l|_{\nu}} u_{a^n|_{a^{-n}\cdot \mu}} s_\mu^* s_\nu^*\\ &=s_\mu s_\nu u_{a^l|_{\nu}} u_{a^n|_\mu} s_\mu^* s_\nu^*\ (\text{as}\ a^{-n}\cdot \mu =\mu)\\ &=(s_\nu u_{a^n} s_\nu^*)(s_\mu u_{a^l} s_\mu^*). \end{align*}

It now follows from lemma 4.15 that ${\mathcal{F}}$ is commutative.\hfill▪

Proof. Similar to [Reference Takeda48], it suffices to show $x\in \operatorname{Ran} \Phi_{\mathbf p}\cap{\mathcal{F}}'$ has the given form. By the Cuntz–Krieger relation, one could assume that $x=s_\mu A s_\nu^*$ with $d(\mu)-d(\nu)=\mathbf p$ and $A\in \mathrm{C}^*(u_a)$. One could further assume that $A=f(u_a)$, where $f(z)=\sum_{i=1}^{n} \lambda_{i} z^{M_i}\ne 0$ for some $0\ne \lambda_i\in {\mathbb{C}}$ and $M_i\in {\mathbb{Z}}$.

Also, it is clear that $s_\mu s_\nu= s_\nu s_\mu$ and $\ s_\mu s_\nu^*= s_\nu^* s_\mu$ for all $\mu, \nu\in \Lambda_{\mathbb{ 1}_{\mathsf k}}$. Then, for all $N\in {\mathbb{Z}}$ and $\omega\in \Lambda_{\mathbb{ 1}_{\mathsf k}}$, one has

\begin{align*} &\ s_\mu A s_{\nu}^*s_{\omega}u_{a^N} s_{\omega}^* - s_{\omega} u_{a^N} s_{\omega}^*s_\mu A s_{\nu}^*\\ =& \ s_\mu \left( \sum_{i=1}^{n} \lambda_i u_{a^{M_i}} \right) s_\omega s_{\nu}^* u_{a^N} s_{\omega}^* - s_{\omega} u_{a^N} s_\mu s_{\omega}^* \left( \sum_{i=1}^{n} \lambda_i u_{a^{M_{i}}} \right) s_{\nu}^* \\ = &\ s_\mu s_\omega \left( \sum_{i=1}^{n} \lambda_{i} u_{a^{M_{i}\mathfrak m^{d(\omega)}}} \right) u_{a^{N\mathfrak m^{d(\nu)}}} s_{\nu}^* s_{\omega}^* - s_{\omega} s_\mu u_{a^{N{\mathbf{m}}^{d(\mu)}}} \left( \sum_{i=1}^{n} \lambda_{i} u_{a^{M_{i} \mathfrak m^{d(\omega)}}} \right) s_{\omega}^* s_{\nu}^* \\ = &\ s_{\mu}s_{\omega} \left( \sum_{i=1}^{n} \lambda_{i} u_{a^{M_{i}\mathfrak m^{d(\omega)}}} \right) \left(u_{a^{N\mathfrak m^{d(\nu)}}} - u_{a^{N{\mathfrak m}^{d(\mu)}}}\right) s_{\omega}^* s_{\nu}^*\\ = &\ s_{\mu}s_{\omega} f\big(u_{a^{\mathfrak m^{d(\omega)}}} \big)\left(u_{a^{N\mathfrak m^{d(\nu)}}} - u_{a^{N{\mathfrak m}^{d(\mu)}}}\right) s_{\omega}^* s_{\nu}^*. \end{align*}

After identifying $\mathrm{C}^*(a)$ with ${\mathrm{C}}({\mathbb{T}})$ (see proposition 2.7), the above is equal to 0 iff

\begin{equation*} f\big(z^{\mathfrak m^{d(\omega)}} \big)\left(z^{N\mathfrak m^{d(\nu)}} - z^{N{\mathfrak m}^{d(\mu)}}\right)=0, \end{equation*}

and therefore, if and only if $\mathfrak m^{d(\mu)}=\mathfrak m^{d(\nu)}$.\hfill▪

Proof. We first show that ${\mathcal{F}}'$ is abelian. For this, let $A:=s_\mu u_{a^M} s_{\nu}^*$ and $B:=s_\alpha u_{a^N} s_\beta ^*$ be two standard generators in ${\mathcal{F}}'$. Then

\begin{align*} AB=s_\mu u_{a^M} s_{\nu}^*s_\alpha u_{a^N} s_\beta^* &=s_\mu u_{a^M} s_\alpha s_{\nu}^* u_{a^N} s_\beta^* =s_\mu s_\alpha u_{a^{M{\mathbf{n}}^{d(\alpha)}}}u_{a^{N{\mathbf{m}}^{d(\nu)}}} s_{\nu}^*s_\beta^*, \\ BA=s_\alpha u_{a^N} s_\beta^* s_\mu u_{a^M} s_{\nu}^* &=s_\alpha u_{a^N} s_\mu s_{\beta}^* u_{a^M} s_\nu^* =s_\mu s_\alpha u_{a^{N{\mathbf{n}}^{d(\mu)}}}u_{a^{M{\mathbf{m}}^{d(\beta)}}} s_{\nu}^*s_\beta^*. \end{align*}

But $\mathfrak m^{d(\mu)}=\mathfrak m^{d(\nu)}$ and $\mathfrak m^{d(\alpha)}=\mathfrak m^{d(\beta)}$ as $A, B\in {\mathcal{F}}'$. Thus AB = BA and so ${\mathcal{F}}'$ is abelian.

Now we show ${\mathcal{F}}'$ is a MASA. Let $s_\alpha u_{a^N} s_{\beta}^*\in {\mathcal{F}}'$ and $s_\mu A s_{\nu}^*\in \operatorname{Ran}\Phi_{\mathbf p}\cap {\mathcal{F}}^{\prime\prime}$ with $A\in \mathrm{C}^*(u_a)$. Similar to the proof of lemma 4.17, we have for all $\mu,\nu\in\Lambda_{\mathbb{ 1}_{\mathsf k}}$ and $N\in {\mathbb{Z}}$

\begin{align*} &\ s_\mu A s_{\nu}^*s_{\alpha} u_{a^N} s_{\beta}^* -s_{\alpha} u_{a^N} s_{\beta}^*s_\mu A s_{\nu}^*\\ =& \ s_\mu \left( \sum_{i=1}^{n} \lambda_{i} u_{a^{M_{i}}} \right) s_\alpha s_{\nu}^* u_{a^N} s_{\beta}^* - s_{\alpha} u_{a^N} s_\mu s_{\beta}^* \left( \sum_{i=1}^{n} \lambda_{i} u_{a^{M_{i}}} \right) s_{\nu}^* \\ =&\ s_\mu s_\alpha \left( \sum_{i=1}^{n} \lambda_{i} u_{a^{M_{i}\mathfrak m^{d(\alpha)}}} \right) u_{a^{N\mathfrak m^{d(\nu)}}} s_{\nu}^* s_{\beta}^* - s_{\alpha} s_\mu u_{a^{N\mathfrak m^{d(\mu)}}} \left( \sum_{i=1}^{n} \lambda_{i} u_{a^{M_{i} \mathfrak m^{d(\beta)}}} \right) s_{\beta}^* s_{\nu}^* \\ =&s_{\alpha} s_\mu \left[\left( \sum_{i=1}^{n} \lambda_{i} u_{a^{M_{i}\mathfrak m^{d(\alpha)}}} \right) u_{a^{N{\mathbf{m}}^{d(\nu)}}} - u_{a^{N\mathfrak m^{d(\mu)}}} \left( \sum_{i=1}^{n} \lambda_{i} u_{a^{M_{i} \mathfrak m^{d(\beta)}}} \right)\right] s_{\beta}^* s_{\nu}^*. \end{align*}

Identify $\mathrm{C}^*(u_a)$ with ${\mathrm{C}}({\mathbb{T}})$ and notice $ m^{d(\alpha)}= m^{d(\beta)}$. Then the above is equal to 0, iff

\begin{equation*} f\big(z^{\mathfrak m^{d(\alpha)}}\big)\left(z^{N\mathfrak m^{d(\nu)}} - z^{N\mathfrak m^{d(\mu)}}\right)=0 \iff \mathfrak m^{d(\mu)}=\mathfrak m^{d(\nu)}. \end{equation*}

Hence $s_\mu A s_{\nu}^*\in {\mathcal{F}}'$ and therefore ${\mathcal{F}}'={\mathcal{F}}^{\prime\prime}$.\hfill▪

Now suppose that $\{|m_i|: 1\le i\le k\}$ is multiplicatively independent. Then ${\mathcal{F}}'={\mathcal{F}}$, which is also the diagonal subalgebra of ${\mathcal{O}}_{\Lambda(\mathbb{ 1}_{\mathsf k},\mathfrak m)}$. So there is a conditional expectation Φ from ${\mathcal{O}}_{\Lambda(\mathbb{ 1}_{\mathsf k},\mathfrak m)}$ onto ${\mathcal{F}}$. Therefore ${\mathcal{F}}$ is a Cartan subalgebra of ${\mathcal{O}}_{\Lambda(\mathbb{ 1}_{\mathsf k},\mathfrak m)}$.

Suppose that $m_i \gt 0$ $(1\le i\le \mathsf k$). For convenience, we use the convention: If $a|_\mu=a^n$, then $t^{\ln a|_\mu}:=t^{\ln n}$. Thus

\begin{align*} \ln a|_{\mu\nu} =\ln \mathfrak m^{d(\mu\nu)} =\ln \mathfrak m^{d(\mu)}+\ln\mathfrak m^{d(\nu)}=\ln a|_\mu+\ln a|_\nu. \end{align*}

Therefore, we obtain action α of ${\mathbb{T}}$ on ${\mathcal{O}}_{\Lambda(\mathbb{ 1}_{\mathsf k},\mathfrak m)}$ as follows:

\begin{equation*} \alpha_t(s_\mu)=t^{\ln a|_\mu} s_\mu, \ \alpha_t(u_a)=u_a \text{for all }\mu \in \Lambda_{\mathbb{ 1}_{\mathsf k}}\text{and }t\in {\mathbb{T}}. \end{equation*}

Define

\begin{equation*} \Psi(x):=\int_{{\mathbb{T}}} \alpha_t(x) d t\quad\text{for all}\quad x\in {\mathcal{O}}_{\Lambda(\mathbb{ 1},\mathfrak m)}. \end{equation*}

Proof. It remains to show that ${\mathcal{F}}'$ is regular. For this, let $A:=s_\mu u_a^M s_{\nu}^*\in {\mathcal{F}}'$ and $B:=s_\alpha u_a^N s_\beta ^*\in {\mathcal{O}}_{\Lambda(\mathbb{ 1}_{\mathsf k},\mathfrak m)}$. Then

\begin{align*} B^*AB &=s_\beta u_{a^{-N}} s_\alpha^* s_\mu u_{a^M} s_{\nu}^*s_\alpha u_{a^N} s_\beta^* \\ &=s_\beta u_{a^{-N}} s_\mu s_\alpha^* u_{a^M} s_\alpha s_{\nu}^* u_{a^N} s_\beta^* \\ &=s_\beta s_\mu u_{a^{-N\mathfrak m^{d(\mu)}}} u_{a^{M\mathfrak m^{d(\alpha)}}} s_\alpha^* s_\alpha s_{\nu}^* u_{a^N} s_\beta^* \\ &=s_\beta s_\mu u_{a^{-N\mathfrak m^{d(\mu)}}} u_{a^{M\mathfrak m^{d(\alpha)}}} u_{a^{N\mathfrak m^{d(\nu)}}} s_{\nu}^* s_\beta^* \\ &=s_\beta s_\mu u_{a^{M\mathfrak m^{d(\alpha)}}} s_\nu^* s_\beta^* \text{(as}\ \mathfrak m^{d(\mu)}=\mathfrak m^{d(\nu)}\text{)}\\ &\in {\mathcal{F}}'. \end{align*}

Therefore ${\mathcal{F}}'$ is regular.

Let G be a discrete (countable) group. A subgroup $S\subseteq G$ is called immediately centralizing if, for every $g\in G$, we either have $\{xgx^{-1}: x\in S\}=\{g\}$ or $\{xgx^{-1}: x\in S\}$ is infinite. This definition is slightly different from the one used in [Reference Duwenig, Gillaspy, Norton, Reznikoff and Wright22] but mentioned in [Reference Duwenig20] and [Reference Duwenig, Gillaspy and Norton21]. Thanks to Anna Duwenig and Rachael Norton for some discussion.

Proof. Let $G:=\langle a, {\mathbf{x}}_i: a{\mathbf{x}}_i={\mathbf{x}}_i a^{m_i}, 1\le i\le \mathsf k\rangle$. Then it it easy to see that ${\mathcal{O}}_{\Lambda(\mathbb{ 1}_{\mathsf k},\mathfrak m)}\cong \mathrm{C}^*(G)$ via $u_a\mapsto a$ and $s_{{\mathbf{x}}_i} \mapsto {\mathbf{x}}_i$. By [Reference Li and Yang34], ${\mathcal{O}}_{\Lambda(\mathbb{ 1}_{\mathsf k},\mathfrak m)}$ is amenable and so is $\mathrm{C}^*(G)$. Thus ${\mathcal{O}}_{\Lambda(\mathbb{ 1}_{\mathsf k},\mathfrak m)}\cong \mathrm{C}^*(G)\cong \mathrm{C}^*_{\text r}(G)$.

Let $S:=\{\mu a^n \nu^{-1}: \mathfrak m^{d(\mu)}=\mathfrak m^{d(\nu)}, n\in {\mathbb{Z}}\}$. Similar to the proof of theorem 4.21, one can easily show that S is a normal subgroup of G. Also, analogous to the proof of lemma 4.18, one can show that, for $\alpha a^k \beta^{-1}\in G$, if the set $\{(\mu a^n \nu^{-1})(\alpha a^k \beta^{-1})(\nu a^{-n} \mu^{-1}): \mu, \nu \in \Lambda(\mathbb{ 1}_{\mathsf k},\mathfrak m), n\in {\mathbb{Z}}\}$ is not a singleton, then it has to be infinite. Indeed,

\begin{align*} (\mu a^n \nu^{-1})(\alpha a^k \beta^{-1})(\nu a^{-n} \mu^{-1}) &= (\mu' a^{n'} \nu'^{-1})(\alpha a^k \beta^{-1})(\nu' a^{-n'} \mu'^{-1})\\ &\iff n \mathfrak{m}^{d(\mu')}=n' \mathfrak{m}^{d(\mu)}. \end{align*}

Hence S is immediately centralizing. By [Reference Duwenig, Gillaspy and Norton21, Theorem 3.1], $\mathrm{C}^*_{\text r}(S)$ is Cartan in $\mathrm{C}^*_{\text r}(G)$. Therefore ${\mathcal{F}}'$ is Cartan in ${\mathcal{O}}_{\Lambda(\mathbb{ 1}_{\mathsf k},\mathfrak m)}$.

4.5. The spectrum of ${\mathcal{F}}$

We end this paper by computing the spectrum of ${\mathcal{F}}$ to connect with Furstenberg’s $\times p, \times q$ conjecture in the viewpoint of [Reference Brownlowe, Ramagge, Robertson and Whittaker9, Reference Furstenberg25]. Let $1\le p_1, \ldots, p_n\in {\mathbb{N}}$ and

\begin{equation*} \varphi: {\mathbb{T}}\to {\mathbb{T}}, \ z\mapsto z^{p_1\cdots p_n}. \end{equation*}

Then the inverse limit

\begin{equation*} \varprojlim({\mathbb{T}}, \varphi):=\left\{(x_n)_{n\in {\mathbb{N}}}\in \prod_{n\in {\mathbb{N}}} {\mathbb{T}}: x_n =\varphi(x_{n+1}) \text{for all }n \in {\mathbb{N}}\right\} \end{equation*}

is a solenoid, denoted as $S_{p_1\cdots p_n}$.

Let $\mathfrak{e}:={\mathbf{x}}_1\cdots {\mathbf{x}}_{\mathsf k}$, the unique path in $\Lambda_{\mathbb{ 1}_{\mathsf k}}$ of degree $\mathbb{ 1}_{\mathsf k}$. For $n \in {\mathbb{N}}$, we have

\begin{align*} {\mathcal{F}}_{n\mathbb{ 1}_{\mathsf k}}&= \overline{\text{span}}\{s_{\mathfrak{e}^n} u_{a^m} s_{\mathfrak{e}^n}^* | \ m \in {\mathbb{Z}} \},\\ {\mathcal{F}} &= \overline{\text{span}}\{s_{\mathfrak{e}^n} u_{a^m} s_{\mathfrak{e}^n}^* | \ n \in {\mathbb{N}}, m \in {\mathbb{Z}} \}=\overline{\bigcup_{n\in {\mathbb{N}}}{\mathcal{F}}_{n\mathbb{ 1}_{\mathsf k}}}. \end{align*}

Note that for any $n \in {\mathbb{N}}$, $s_{\mathfrak{e}^n} u_a s_{\mathfrak{e}^n}^*$ is a unitary that generates ${\mathcal{F}}_{n\mathbb{ 1}_{\mathsf k}}$. Then one can see that there is an isomorphism $\psi_n:{\mathcal{F}}_{n\mathbb{ 1}_{\mathsf k}} \to {\mathrm{C}}({\mathbb{T}})$.

Let $\mathfrak M:=\prod_{i=1}^{\mathsf k} m_i$.

Proof. Notice that

\begin{equation*} s_{\mathfrak{e}^n} u_{a^p} s_{\mathfrak{e}^n}^* = s_{\mathfrak{e}^n} s_{\mathfrak{e}}s_{\mathfrak{e}}^* u_{a^p} s_{\mathfrak{e}^n}^* = s_{\mathfrak{e}^{n+1}} u_{a^{p\mathfrak M}} s_{\mathfrak{e}^{n+1}}^*. \end{equation*}

This shows that φ_n is indeed an inclusion of C*-algebras. Since the union $\bigcup_{n=0}^{\infty} {\mathcal{F}}_{n\mathbb{ 1}_{\mathsf k}}$ is dense in ${\mathcal{F}}$ (actually they are equal), we have that ${\mathcal{F}}$ is isomorphic to the direct limit of $({\mathcal{F}}_{n\mathbb{ 1}_{\mathsf k}},\varphi_n)$ (see [Reference Li and Yang37, Remark 6.1.3]).

Due to [Reference Scarparo45, Theorem 2], the spectrum of ${\mathcal{F}}$ is the projective limit of the spectra of the subalgebras ${\mathcal{F}}_{n\mathbb{ 1}_{\mathsf k}}$, with the maps $\phi_{n+1}:\widehat{{\mathcal{F}}_{(n+1)\mathbb{ 1}_{\mathsf k}}} \rightarrow \widehat{{\mathcal{F}}_{n\mathbb{ 1}_{\mathsf k}}}$ which induce the maps φ_n. However, we will work with an isomorphic direct system in order to make things more concrete. Observe that we have an isomorphism ψ_n between ${\mathcal{F}}_{n\mathbb{ 1}_{\mathsf k}}$ and ${\mathrm{C}}({\mathbb{T}})$ that sends the element $s_{\mathfrak{e}^n} a s_{\mathfrak{e}^n}^* \in {\mathcal{F}}_n$ to the function $f(z) = z$ in ${\mathrm{C}}({\mathbb{T}})$, which we will just denote by z (note that z is a unitary element that generates ${\mathrm{C}}({\mathbb{T}})$ ). Let $\varphi'_{n}: {\mathrm{C}}({\mathbb{T}}) \rightarrow {\mathrm{C}}({\mathbb{T}})$ be the map defined by sending the function z to $z^{\mathfrak M}$. Then by a direct calculation the following diagram commutes:

By [Reference Scarparo45, Proposition 2], ${\mathcal{F}}$ is then isomorphic to the direct limit of the system $({\mathrm{C}}({\mathbb{T}}), \varphi'_{n,n+1})$. We now observe that homeomorphism $\rho:{\mathbb{T}} \rightarrow {\mathbb{T}}$ defined by $\rho(z) = z^{\mathfrak M}$ induces the maps $\varphi'_n:{\mathrm{C}}({\mathbb{T}})\rightarrow {\mathrm{C}}({\mathbb{T}})$, and so the spectrum ${\mathcal{F}}$ is homeomorphic to the projective limit of

\begin{equation*} {\mathbb{T}} \xleftarrow{\rho} {\mathbb{T}} \xleftarrow{\rho} {\mathbb{T}} \xleftarrow{\rho} {\mathbb{T}} \xleftarrow{\rho} \ldots \end{equation*}

which is precisely $S_{\mathfrak M}$.\hfill▪

Remark 4.25. One can show that ${\mathcal{O}}_{\Lambda(\mathbb{ 1}_{\mathsf k},\mathfrak m)}\cong {\mathcal{F}} \lt imes {\mathbb{Z}}^{\mathsf k}\cong {\mathrm{C}}(S_{\mathfrak M}) \lt imes {\mathbb{Z}}^{\mathsf k}$. In fact, the action of ${\mathbb{Z}}^{\mathsf k}$ on ${\mathcal{F}}$ is given by

\begin{equation*} \alpha: {\mathbb{Z}}^{\mathsf k}\to \operatorname{Aut}{\mathcal{F}}, \ \alpha_{\mathbf n}(A)= s_\nu A s_\nu^*, \end{equation*}

where ν is the unique path in $\Lambda_{\mathbb{ 1}_{\mathsf k}}$ of degree $\mathbf n$.