C*-Algebras of Right LCM One-Relator Monoids and Artin–Tits Monoids of Finite Type

Abstract

We study C*-algebras generated by left regular representations of right LCM one-relator monoids and Artin–Tits monoids of finite type. We obtain structural results concerning nuclearity, ideal structure and pure infiniteness. Moreover, we compute K-theory. Based on our K-theory results, we develop a new way of computing K-theory for certain group C*-algebras and crossed products.

1 Introduction

Ever since the work in [9, 13], C*-algebras generated by isometries have provided interesting examples in the theory of general C*-algebras. In particular, C*-algebras generated by left regular representations of left-cancellative semigroups, also called semigroup C*-algebras, form a natural class of C*-algebras which have been studied for several types of semigroups (see [18] and the references therein). This construction has attracted attention because of connections to several topics such as index theory, representation theory, amenability, and non-selfadjoint operator algebras. Recently, it was discovered in [15, 31,32,33] that there is even a connection to number theory. This observation has led to a renewed interest in the subject. Apart from these examples from number theory, there is another natural class of examples coming from group theory, for instance semigroups given by generators and relations.

This present paper makes a contribution towards a better understanding of the C*-algebras attached to this second class of monoids. More precisely, we study semigroup C*-algebras of two classes of such monoids: one-relator monoids and Artin–Tits monoids of finite type. Our motivating example is the classical Braid monoid $B_n^+$ which sits inside the Braid group $B_n$. Geometrically speaking, elements of $B_n^+$ can be thought of as braids in which the crossings are allowed to go in one direction only. The (reduced) semigroup C*-algebra of $B_n^+$ turns out to be an extension of the reduced group C*-algebra of $B_n$ by another C*-algebra which is very similar to a graph C*-algebra [45] in the case $n=3$ and which is not nuclear in the case $n \ge 4$. Our search for other classes of monoids whose C*-algebras have a similar structure as the one of $B_3^+$ has led us to one-relator monoids, while monoids whose C*-algebras behave like the ones of $B_n^+$, for $n \ge 4$, are given by more general Artin–Tits monoids (see [6, 24, 47]). In this paper, we mainly focus on structural results and K-theory computations for these classes of semigroup C*-algebras. The latter turn out to be interesting because they lead to a new way of computing K-theory for certain group C*-algebras and crossed products.

Let us now describe our main results in more detail. For a class of cancellative right LCM one-relator monoids as specified in Sects. 3 and 4, we show that there is a big difference between the reversible and the non-reversible cases. In the non-reversible case, we show that the boundary quotient (in the sense of [18, § 5.7]) of our semigroup C*-algebra is purely infinite simple (see Corollary 3.2, where this is proven in even greater generality). It follows that the semigroup C*-algebra is the extension of a purely infinite simple C*-algebra by the algebra of compact operators if the underlying monoid is finitely generated, and it is purely infinite simple itself if the monoid is not finitely generated (see Corollary 3.5). Moreover, this boundary quotient can be identified with a graph C*-algebra [45] in special cases, so that we can further deduce that our semigroup C*-algebra is nuclear (see Corollary 3.10).

In the reversible case, the boundary quotient is given by the reduced C*-algebra of the one-relator group given by the same presentation as our one-relator monoid. This means that the boundary quotient is almost never nuclear since most one-relator groups are not amenable. However, we show that in special cases, the kernel of the canonical projection map onto the boundary quotient is a Toeplitz graph C*-algebra, in particular nuclear (see Theorem 4.3).

The main advantage of our new approach is that we are able to identify general conditions on the presentations defining our monoids which allow us to deduce the above-mentioned structural results. This is in contrast to previous work, which focused on case-by-case studies of classes of monoids defined by generators and relations, where the methods had to be adapted to the defining relations.

Our analysis of one-relator monoids prompts the question whether similar results still hold in the case of more relations. We show that in general, this is not the case. For C*-algebras attached to Artin–Tits monoids of finite type, which are not one-relator monoids, not only does the boundary quotient fail to be nuclear, but even the kernel of the canonical projection onto the boundary quotient is typically not nuclear (see Proposition 4.15). However, we show that this kernel is still purely infinite simple for irreducible Artin–Tits monoids with at least three generators (see Theorem 4.39). Our results complement the work in [10, 11, 25] on C*-algebras of right-angled Artin–Tits monoids.

Finally, we use our analysis of the ideal corresponding to the boundary quotient to provide general tools for K-theory computations. We focus on dihedral Artin–Tits groups and torus knot groups. In each of these cases, we construct two six-term exact sequences in K-theory. The first one allows to compute K-theory for the ideal corresponding to the boundary quotient, while the second one requires K-theory of the ideal as an input and determines K-theory of the reduced group C*-algebras of dihedral Artin–Tits groups and torus knot groups. All this works for arbitrary coefficients and therefore allows for K-theory computations for crossed products (see Theorem 5.6). We then present explicit K-theory computations for a variety of examples, including group C*-algebras of dihedral Artin–Tits groups, torus knot groups, the braid group $B_4$ and a semidirect product group arising from Artin’s representation of braid groups (see Sect. ).

The main ingredient for our computations is a K-theory formula for crossed products in the presence of the independence condition (see [16, 17]). Therefore, our K-theory results point towards the possibility of using K-theory formulas for semigroup C*-algebras and closely related crossed products as in [16, 17] to compute K-theory for group C*-algebras and crossed products. Moreover, since we want to allow arbitrary coefficients, we introduce graph C*-algebras twisted by coefficient algebras (a special case of this construction appears in [14]). This construction might be of independent interest.

We would like to thank Nathan Brownlowe and Dave Robertson for inviting us to the workshop “Interactions Between Semigroups and Operator Algebras” in Newcastle (Australia), where this project has been initiated. Moreover, we thank the referees for many helpful comments which improved our paper considerably.

2 Preliminaries

Let us start by reviewing some basics concerning semigroups and their C*-algebras.

2.1 Monoids

Semigroups with an identity are called monoids in this paper. Although it is often not strictly necessary, we will focus on monoids throughout, just to keep things simple and because all our examples do have an identity.

2.1.1 Monoids defined by presentations

Let $\Sigma$ be a countable set, $\Sigma ^*$ the free monoid generated by $\Sigma$ (also viewed as the set of finite words in $\Sigma$) and R a subset of $\Sigma ^* \times \Sigma ^*$. We call $(\Sigma ,R)$ a presentation, where $\Sigma$ is the set of generators and R is the set of relations. Words $u \in \Sigma ^*$ for which (u, v) or (v, u) lies in R are called relators. Given a presentation $(\Sigma ,R)$, we form the monoid $P = \left\langle \Sigma \, \vert \, R \right\rangle ^+$ generated by $\Sigma$ subject to the relations $u = v$ for all $(u,v) \in R$ (see [8, § 1.12]). Note that countability of $\Sigma$ is not essential; this assumption is included only to guarantee that our semigroup C*-algebras are separable.

Given $x \in \Sigma ^*$, we also denote by x the element in P represented by x. Given $x, y \in \Sigma ^*$, we write $x \equiv y$ if x and y coincide in $\Sigma ^*$, i.e., x is the same word as y, and $x = y$ if x and y represent the same element in P. We write $\ell ^*$ for the length of a word in $\Sigma ^*$ and $\ell (x) \mathrel {:=}\min \left\{ \ell ^*(w) \text {: }w \in \Sigma ^*, \, w = x \right\}$ for the word length of $x \in P$ with respect to $\Sigma$. Given $\sigma \in \Sigma$ and $x \in \Sigma ^*$, $\ell _{\sigma }(x)$ counts how many times $\sigma$ appears in x. We write $\varepsilon$ for the empty word in $\Sigma ^*$.

Recall how $P = \left\langle \Sigma \, \vert \, R \right\rangle ^+$ is constructed: For two words $w, w' \in \Sigma ^*$, we write $w \equiv _R w'$ if $w \equiv w'$ or there exist $a, u, v, z \in \Sigma ^*$ with $(u,v) \in R$ or $(v,u) \in R$ such that $w \equiv a u z$ and $w' \equiv a v z$. Two words $x, y \in \Sigma ^*$ represent the same element in P if and only if there is a finite sequence $w_1, \dotsc , w_s \in \Sigma ^*$ with $w_1 \equiv x$, $w_s \equiv y$, and for all $1 \le i \le s-1$, we have $w_i \equiv _R w_{i+1}$.

2.1.2 Standing assumptions on our presentations

All our presentations will satisfy that for all $(u,v) \in R$, we have $u \ne \varepsilon \ne v$. It then immediately follows from the construction of $P = \left\langle \Sigma \, \vert \, R \right\rangle ^+$ that P does not contain any left- or right-invertible elements other than the identity. In particular, the group of units $P^*$ in P is trivial, i.e., $P^* = \left\{ e \right\}$, where e is the identity in P.

In the following, we will also always exclude the (degenerate) case that a generator $\sigma \in \Sigma$ is redundant, i.e., we always assume that for every $\sigma \in \Sigma$ and $w \in (\Sigma {\setminus } \left\{ \sigma \right\} )^*$ that $\sigma \ne w$. Otherwise, if $\sigma \in \Sigma$ satisfies $\sigma = w$ for some $w \in (\Sigma {\setminus } \left\{ \sigma \right\} )^*$, then we could always give another presentation for P, $P \cong \left\langle \Sigma {\setminus } \left\{ \sigma \right\} \, \vert \, R' \right\rangle ^+$, where $R'$ is obtained from R by replacing every $(u,v) \in R$ by $(u',v')$, where we obtain $u'$ from u and $v'$ from v by replacing every $\sigma$ in u and v by w.

Moreover, we will always assume (without loss of generality) that R does not contain (u, u) for any $u \in \Sigma ^*$.

Note that under these standing assumptions, $P = \left\langle \Sigma \, \vert \, R \right\rangle ^+$ is finitely generated if and only if $|\Sigma | < \infty$.

2.1.3 One-relator monoids

We will mainly focus on the case of one-relator monoids, i.e., monoids of the form $P = \left\langle \Sigma \, \vert \, R \right\rangle ^+$ where R consists of a single pair $(u,v) \in \Sigma ^* \times \Sigma ^*$. This is (arguably) the most basic case. Moreover, in examples, we actually see very different phenomena when we drop this assumption (see Sect. 4.2).

We often write $(\Sigma , u=v)$ for $(\Sigma , R)$ and $\left\langle \Sigma \, \vert \, u = v \right\rangle ^+$ for $\left\langle \Sigma \, \vert \, R \right\rangle ^+$. Let $G = \left\langle \Sigma \, \vert \, u = v \right\rangle$ be the group generated by $\Sigma$ subject to the relation $u = v$. Let us assume that the first letter of u is not equal to the first letter of v and that the last letter of u is not equal to the last letter of v. Then it was proven in [1] that the canonical map $P \rightarrow G$ induced by the identity on $\Sigma$ is injective. In particular, P is cancellative (i.e., left and right cancellative). Conversely, since we assume that $u \not \equiv v$, if P is cancellative, then the first and last letters of u and v must differ.

Therefore, as we do want cancellation to hold, whenever we consider one-relator monoids of the form $\left\langle \Sigma \, \vert \, u = v \right\rangle ^+$, we will assume that the first letter of u is not equal to the first letter of v and that the last letter of u is not equal to the last letter of v.

Remark 2.1

$G = \left\langle \Sigma \, \vert \, u = v \right\rangle$ for $u, v \in \Sigma ^*$ is a one-relator group. If the first and last letters of u and v differ, then it follows from [37, Theorem 4.12] that G is torsion-free. Moreover, G is amenable if and only if G is cyclic or $G \cong \left\langle a,b \, \vert \, ab = b^ka \right\rangle$ for some $0 \ne k \in \mathbb {Z}$ (see for instance [7]).

2.1.4 The LCM property

We will assume throughout Sects. 3–5 that all our monoids P are right LCM, i.e., given $p, q \in P$, we have $pP \cap qP = \emptyset$ or there exists $r \in P$ with $pP \cap qP = rP$.

Applying the notion of right reversing in the sense of [21] to our presentation $(\Sigma , u=v)$, let us present a sufficient condition for the right LCM property:

Let us introduce another copy of $\Sigma$ by forming $\Sigma \cup \Sigma ^{-1} = \left\{ \sigma \text {: }\sigma \in \Sigma \right\} \cup \left\{ \sigma ^{-1} \text {: }\sigma \in \Sigma \right\}$. For $x, y \in (\Sigma \cup \Sigma ^{-1})^*$, we write $x \curvearrowright _r y$ if there exist $w_1, w_2, \dotsc , w_j \in (\Sigma \cup \Sigma ^{-1})^*$ with $x \equiv w_1$, $y \equiv w_j$, and for all $1 \le i \le j-1$, we have $w_i \equiv \alpha \sigma ^{-1} \tau \zeta$ for $\sigma , \tau \in \Sigma$ and $\alpha , \zeta \in (\Sigma \cup \Sigma ^{-1})^*$, and $w_{i+1} \equiv \alpha \zeta$ if $\sigma \equiv \tau$, and $w_{i+1} \equiv \alpha s t^{-1} \zeta$ if $\sigma s \equiv u$ and $\tau t \equiv v$. Then, following [21, Definition 3.1], we say that $(\Sigma , u=v)$ satisfies the strong $\varvec{r}$-cube condition on $\Sigma$ if for all $\sigma , \tau , \upsilon \in \Sigma$, $\sigma ^{-1} \tau \tau ^{-1} \upsilon \curvearrowright _r xy^{-1}$ for $x, y \in \Sigma ^*$ implies that $(\sigma x)^{-1} (\upsilon y) \curvearrowright _r \varepsilon$. Note that for a word $w = \sigma _1 \sigma _2 \cdots \sigma _{l-1} \sigma _l \in \Sigma ^*$, we write $w^{-1} = \sigma _l^{-1} \sigma _{l-1}^{-1} \cdots \sigma _2^{-1} \sigma _1^{-1} \in (\Sigma ^{-1})^*$.

Now, in our case of the presentation $(\Sigma , u=v)$, it is easy to verify the strong $\varvec{r}$-cube condition on $\Sigma$: If $\sigma \not \equiv \tau$ and $\sigma , \tau$ are not the first letters of u and v, then we cannot have $\sigma ^{-1} \tau \tau ^{-1} \upsilon \curvearrowright _r xy^{-1}$ for some $x, y \in \Sigma ^*$. Similarly, if $\tau \not \equiv \upsilon$ and $\tau , \upsilon$ are not the first letters of u and v, then we cannot have $\sigma ^{-1} \tau \tau ^{-1} \upsilon \curvearrowright _r xy^{-1}$ for some $x, y \in \Sigma ^*$. If $\sigma \equiv \tau \equiv \upsilon$, then $x \equiv y \equiv \varepsilon$ and the strong $\varvec{r}$-cube condition on $\Sigma$ is obviously satisfied. If $\sigma \equiv \tau$ and $u \equiv \tau x$, $v \equiv \upsilon y$, then $\sigma ^{-1} \tau \tau ^{-1} \upsilon \equiv \tau ^{-1} \tau \tau ^{-1} \upsilon \curvearrowright _r xy^{-1}$, and $(\tau x)^{-1} (\upsilon y) \curvearrowright _r \varepsilon$. The case $u \equiv \sigma x$, $v \equiv \tau y$ and $\tau \equiv \upsilon$ is analogous. Finally, if $\sigma \equiv \upsilon$ and $u \equiv \sigma s \equiv \upsilon s$, $v \equiv \tau t$, then $\sigma ^{-1} \tau \tau ^{-1} \upsilon \curvearrowright _r s t^{-1} t s^{-1} \curvearrowright _r ss^{-1}$, so that $x \equiv s$, $y \equiv s$, and $(\sigma s)^{-1} (\upsilon s) \equiv (\sigma s)^{-1} (\sigma s) \curvearrowright _r \varepsilon$. This shows the strong $\varvec{r}$-cube condition on $\Sigma$.

In order to deduce the strong $\varvec{r}$-cube condition (on $\Sigma ^*$), we need to check whether $(\Sigma , u=v)$ is $\varvec{r}$-homogeneous in the sense of [21, Definition 4.1]. This is for instance the case if $\ell ^*(u) = \ell ^*(v)$ (for we can take $\lambda = \ell ^*$ in [21, Definition 4.1]). $(\Sigma , u=v)$ is also $\varvec{r}$-homogeneous if $\ell ^*(u) < \ell ^*(v)$ and there exists $\sigma \in \Sigma$ with $\ell _{\sigma }(u) > \ell _{\sigma }(v)$. In that case, let $\delta \mathrel {:=}\ell ^*(v) - \ell _{\sigma }(v) - (\ell ^*(u) - \ell _{\sigma }(u))$. Then $0< \ell _{\sigma }(u) - \ell _{\sigma }(v) < \delta$. Let $\eta$ and $\zeta$ be positive integers with $\delta \eta = \mathrm{lcm}(\delta , \ell _{\sigma }(u) - \ell _{\sigma }(v)) = (\ell _{\sigma }(u) - \ell _{\sigma }(v)) \zeta$. Then define $\lambda$ as the monoid homomorphism from $\Sigma ^*$ to the non-negative integers by setting $\lambda (\sigma ) = \zeta$ and $\lambda (\tau ) = \eta$ for all $\tau \in \Sigma {\setminus } \left\{ \sigma \right\}$. We then have $\lambda (u) = \zeta \ell _{\sigma }(u) + \eta (\ell ^*(u) - \ell _{\sigma }(u)) = \delta \eta + \zeta \ell _{\sigma }(v) + \eta (\ell ^*(u) - \ell _{\sigma }(u)) = \zeta \ell _{\sigma }(v) + \eta (\delta + \ell ^*(u) - \ell _{\sigma }(u)) = \zeta \ell _{\sigma }(v) + \eta (\ell ^*(v) - \ell _{\sigma }(v)) = \lambda (v)$. Hence $(\Sigma , u=v)$ is $\varvec{r}$-homogeneous. Once we know that $(\Sigma , u=v)$ is $\varvec{r}$-homogeneous, [21, Proposition 4.4] tells us that $(\Sigma , u=v)$ is $\varvec{r}$-complete in the sense of [21, Definition 2.1]. This in turn implies that $P = \left\langle \Sigma , u=v \right\rangle ^+$ is right LCM by Dehornoy [21, Proposition 6.10].

Remark 2.2

Note that there are examples of presentations $(\Sigma , u=v)$ which are not $\varvec{r}$-homogeneous but where $\left\langle \Sigma , u=v \right\rangle ^+$ is still right LCM. For instance, let $\Sigma = \left\{ a,b \right\}$ and $u = b^d a b^c$ for some positive integers c, d, and $v = a$. If we define $x_i \mathrel {:=}a b^{ic}$ for $i = 0, 1, \dotsc$, then $x_i = b^d x_{i+1}$ for all i. Thus $\left\langle \Sigma \, \vert \, u=v \right\rangle ^+$ is not $\varvec{r}$-Noetherian in the sense of [21, Definition 2.6], so that $(\Sigma , u=v)$ cannot be $\varvec{r}$-homogeneous by Dehornoy [21, Proposition 4.3]. However, $\left\langle \Sigma \, \vert \, u=v \right\rangle ^+$ is right LCM by Spielberg [50, Proposition 2.10].

2.1.5 Reversibility

Let us discuss the question when $P = \left\langle \Sigma \, \vert \, u=v \right\rangle ^+$ is left reversible, i.e., for all $p, q \in P$, we have $pP \cap qP \ne \emptyset$. We first observe that if $(\Sigma , u=v)$ is $\varvec{r}$-homogeneous, then by Dehornoy [21, Proposition 6.7], $P = \left\langle \Sigma \, \vert \, u=v \right\rangle ^+$ is left reversible if and only if there exists $\Sigma \subseteq \Sigma ' \subseteq \Sigma ^*$ such that for all $w, x \in \Sigma '$, there exist $y, z \in \Sigma '$ with $(xz)^{-1}(wy) \curvearrowright _r \varepsilon$. In particular, $P = \left\langle \Sigma \, \vert \, u=v \right\rangle ^+$ is left reversible if there exists $\Sigma \subseteq \Sigma ' \subseteq \Sigma ^*$ such that for all $w, x \in \Sigma '$, there exist $y, z \in \Sigma '$ with $x^{-1}w \curvearrowright _r zy^{-1}$ (see [21, Remark 6.9]).

If $|\Sigma | \ge 3$, then P cannot be left reversible: Take $\sigma \in \Sigma$ such that $\sigma$ is not the first letter of u and also not the first letter of v. Then for every $\tau \in \Sigma {\setminus } \left\{ \sigma \right\}$, we must have $\sigma P \cap \tau P = \emptyset$ because we cannot find $t, w \in \Sigma ^*$ with $tuw \equiv tvw$ such that the first letter of tuw is $\sigma$ and the first letter of tvw is $\tau$ (or vice versa).

Moreover, let us present examples of presentations $(\left\{ a,b \right\} , u=v)$ which do not give rise to left reversible monoids. Define $\mathrm{OVL}(v) = \left\{ x \in \left\{ a,b \right\} ^* \text {: }v \equiv xy \equiv wx \right\}$ for some $\varepsilon \ne w, y \in \left\{ a,b \right\} ^*$. Assume that $\mathrm{OVL}(v) = \left\{ \varepsilon \right\}$. Further suppose that $\ell _a(u) < \ell _a(v)$ or $\ell _b(u) < \ell _b (v)$. The first condition implies that $(\left\{ a,b \right\} , u=v)$ satisfies the Church–Rosser condition (by Trubitsyn [51]), while the second condition ensures that $(\left\{ a,b \right\} , u=v)$ is noetherian. (The reader may consult [4, § 1.1] for explanations of these notions.) This implies that given $w \in \left\{ a,b \right\} ^*$, we can find a unique irreducible word $x \in \left\{ a,b \right\} ^*$ (i.e., x does not contain v as a subword) such that $w = x$ in P, and two irreducible words $x, y \in \left\{ a,b \right\} ^*$ represent the same element in P if and only if $x \equiv y$ (see [4, Theorem 1.1.12]). Now assume that $\mathrm{OVL}(v) = \left\{ \varepsilon \right\}$, and that $\ell _a(u) < \ell _a(v)$ or $\ell _b(u) < \ell _b (v)$. If the first letter of v is a, but $v \not \equiv ab^k$ for some $k \ge 1$, then P is not left reversible: Let $p = b^m$, $q = a b^n$ for some $m, n \ge \ell ^*(v)$. Take $x, y \in \left\{ a,b \right\} ^*$ irreducible. Then px and qy are irreducible, but $px \ne qy$ as $px \not \equiv qy$. Hence $pP \cap qP = \emptyset$.

2.2 Semigroup C*-algebras

Let P be a left cancellative semigroup and $\lambda$ its left regular representation on $\ell ^2 P$. This means that for $p \in P$, $\lambda (p)$ is the isometry $\ell ^2 P \rightarrow \ell ^2 P$ determined by $(\lambda (p)\xi )(px) = \xi (x)$ for all $x \in P$ and $(\lambda (p)\xi )(y) = 0$ if $y \notin pP$. We denote by $C^*_{\lambda }(P)$ the C*-algebra generated by $\lambda (P)$, and call it the (reduced) semigroup C*-algebra of P. Let S be the inverse semigroup of partial isometries on $\ell ^2 P$ generated by $\lambda (P)$. Let $D_{\lambda }(P) \mathrel {:=}C^*(\left\{ s^*s \text {: }s \in S \right\}$. If P embeds into a group G, then we can also describe $D_{\lambda }(P)$ by $D_{\lambda }(P) = C^*_{\lambda }(P) \cap \ell ^{\infty }(P)$. $D_{\lambda }(P)$ is always a commutative C*-algebra, and we define $\Omega _P \mathrel {:=}\mathrm{Spec\,}\left( D_{\lambda }(P) \right)$. We refer the reader to [29, 30] as well as [18, § 5] for more details about general semigroup C*-algebras.

2.2.1 Description as groupoid C*-algebras

If P is a right LCM monoid, then $D_{\lambda }(P) = C^*(\left\{ 1_{pP} \text {: }p \in P \right\} ) \subseteq \ell ^{\infty }(P)$. Let $\mathcal J_P \mathrel {:=}\left\{ pP \text {: }p \in P \right\} \cup \left\{ \emptyset \right\}$. $\mathcal J_P$ is a semilattice under intersection because P is right LCM. Points in $\Omega _P$, i.e., characters on $D_{\lambda }(P)$, are in bijection to multiplicative surjective maps $\chi : \, \mathcal J_P \rightarrow \left\{ 0,1 \right\}$ with $\chi (\emptyset ) = 0$ (such a map $\chi$ corresponds to the character $\omega : \, D_{\lambda }(P) \rightarrow \mathbb {C}$ determined by $\omega (1_{pP}) = \chi (pP)$). The topology on $\Omega _P$ corresponds to the topology of point-wise convergence for maps $\mathcal J_P \rightarrow \left\{ 0,1 \right\}$. In the following, we identify $\Omega _P$ with multiplicative surjective maps $\chi : \, \mathcal J_P \rightarrow \left\{ 0,1 \right\}$ as explained above.

If P embeds into a group G, then we can fix such an embedding and view P as a submonoid of G, and consider the partial action $G \curvearrowright \Omega _P$ determined by the partial homeomorphisms $U_{g^{-1}} \rightarrow U_g, \, \chi \mapsto g.\chi$, where $U_{g^{-1}}$ is the subspace of all $\chi \in \Omega _P$ which satisfy $\chi (qP) = 1$ for some $q \in P$ such that $g = pq^{-1}$ for some $p \in P$. For such $\chi$, $g.\chi$ is determined by $(g.\chi )(xP) = \chi (qyP)$ if $xP \cap pP = pyP$ and $(g.\chi )(xP) = 0$ if $xP \cap pP = \emptyset$. Note that $U_{g^{-1}}$ is non-empty if and only if $g = pq^{-1}$ for some $p, q \in P$. As explained in [18, Theorem 5.6.41], the semigroup C*-algebra of P is canonically isomorphic to the reduced crossed product attached to this partial dynamical system $G \curvearrowright \Omega _P$ as well as isomorphic to the reduced C*-algebra of the corresponding partial transformation groupoid, $C^*_{\lambda }(P) \cong C^*_r(G \ltimes \Omega _P)$.

2.2.2 Distinguished subspaces of the character space

For brevity, let us drop indices and write $\mathcal J\mathrel {:=}\mathcal J_P$, $\mathcal J^\times = \mathcal J{\setminus } \left\{ \emptyset \right\}$ and $\Omega \mathrel {:=}\Omega _P$. Let us describe distinguished elements and subspaces of $\Omega$. First of all, every $pP \in \mathcal J$ determines a point $\chi _{pP} \in \Omega$ given by $\chi _{pP}(xP) = 1$ if $pP \subseteq xP$ and $\chi _{pP}(xP) = 0$ if $pP \nsubseteq xP$. This allows us to identify $\mathcal J^\times$ with a subset of $\Omega$. It is easy to see that $\mathcal J^\times$ is dense in $\Omega$. We define $\Omega _{\infty } = \Omega {\setminus } \mathcal J^\times$. Clearly, $\mathcal J^\times$ and $\Omega _{\infty }$ are G-invariant subspaces. In the case where $P^* = \left\{ e \right\}$, we have a bijection $P \cong \mathcal J^\times , \, p \mapsto pP$, so that we will not distinguish between P and $\mathcal J^\times$, and hence we have $\Omega _{\infty } = \Omega {\setminus } P$. Among the points in $\Omega _{\infty }$, we single out those $\chi \in \Omega$ for which $\chi ^{-1}(1)$ is maximal, i.e., whenever $\omega \in \Omega$ satisfies $\omega (xP) = 1$ for all $xP \in \mathcal J$ with $\chi (xP) = 1$, then we must have $\omega = \chi$. We set $\Omega _{\max } \mathrel {:=}\left\{ \chi \in \Omega \text {: }\chi ^{-1}(1) \ \mathrm{is} \ \mathrm{maximal} \right\}$. Note that $\Omega _{\max } \subseteq \Omega _{\infty }$. Moreover, we define $\partial \Omega \mathrel {:=}\overline{\Omega _{\max }}$. Let us now collect a few facts about $\partial \Omega$, which are obtained in [18, § 5.7] in greater generality than needed here. $\partial \Omega$ is the minimal non-empty closed G-invariant subspace of $\Omega$. Moreover, $\partial \Omega$ reduces to a single point (namely $\chi \in \Omega$ given by $\chi (xP) = 1$ for all $x \in P$; we usually denote this $\chi$ by $\infty$) if and only if P is left reversible. If $\partial \Omega$ is not a point, then $G \curvearrowright \partial \Omega$ is purely infinite. Since $\partial \Omega$ is always a closed G-invariant subspace of $\Omega$, we can define a quotient of $C^*_{\lambda }(P)$ by setting $\partial C^*_{\lambda }(P) \mathrel {:=}C^*_r(G \ltimes \partial \Omega )$. $\partial C^*_{\lambda }(P)$ is called the boundary quotient of $C^*_{\lambda }(P)$.

2.2.3 Characters attached to words

Now let us assume that our monoid P is given by a presentation $(\Sigma , R)$ as in Sect. 2.1.1 satisfying the standing assumptions in Sect. 2.1.2. Assume that P is right LCM. Let G be the group given by the same presentation as P, and suppose that $P \rightarrow G$ induced by $\mathrm{id}_{\Sigma }$ is an embedding (this must be the case if P embeds into a group). Since $P^* = \left\{ e \right\}$ we may write P for the subset $\mathcal J$ of $\Omega$. In this setting, let us describe $\Omega _{\infty }$ by infinite words (with letters) in $\Sigma$. We write $\Sigma ^{\infty }$ for the set of all these infinite words. Given $w \in \Sigma ^{\infty }$, define $w_{[1,i]}$ to be the word consisting of the first i letters of w (for $i \ge 1$). Let us write $w_i \mathrel {:=}w_{[1,i]}$. Then define $\chi _w \in \Omega$ by setting $\chi _w(zP) = 1$ if and only if $w_i \in zP$ for some i. It is easy to see that $\chi _w \in \Omega _{\infty }$. Conversely, we have

Lemma 2.3

Every $\chi \in \Omega _{\infty }$ is of the form $\chi _w$ for some $w \in \Sigma ^{\infty }$.

Proof

Given $\chi \in \Omega _{\infty }$, let us enumerate $zP \in \mathcal J$ with $\chi (zP) = 1$, so that $\left\{ zP \text {: }\chi \right\} (zP) = 1 = \left\{ z_1 P, z_2 P, \dotsc \right\}$. Then choose $y_i \in \Sigma ^*$ with $y_i P = z_1 P \cap \cdots \cap z_i P$ and $y_{i+1} \in y_i \Sigma ^*$. Note that all $y_i P$ themselves lie in $\left\{ z_1 P, z_2 P, \dotsc \right\}$. Let $l_i \mathrel {:=}\ell ^*(y_i)$. Since $\chi \in \Omega _{\infty }$, we must have $\lim _{i \rightarrow \infty } l_i = \infty$. So there is a unique infinite word w in $\Sigma$ such that $w_{l_i} = y_i$. We claim that $\chi = \chi _w$. Indeed, given $x \in P$, we have $\chi (xP) = 1$ if and only if $xP \in \left\{ z_1P, z_2P, \dotsc \right\}$ if and only if $y_i \in xP$ for some i if and only if $w_j \in xP$ for some j if and only if $\chi _w(xP) = 1$. $\square$

In this picture with infinite words, the G-action on $\Omega _{\infty }$ is given as follows: $\chi \in \Omega _{\infty }$ lies in $U_{g^{-1}}$ if and only if $\chi = \chi _w$ for some $w \in \Sigma ^{\infty }$ which starts with $q \in \Sigma ^*$, i.e., of the form $w \equiv qx$ for some $x \in \Sigma ^{\infty }$, and $g = pq^{-1}$ for some $p \in \Sigma ^*$, and in this case we have $g.\chi = g.\chi _w = \chi _{px}$.

The topology on $\Omega _{\infty }$ is the subspace topology from $\Omega$, which in turn is determined by the basic open subspaces $U(pP; p_1P, \dotsc , p_k P) \mathrel {:=}\left\{ \chi \in \Omega \text {: }\chi (pP) = 1 \right\}$, $\chi (p_1 P) = \cdots = \chi (p_k P) = 0$, where $p_j P \subsetneq pP$ for all $1 \le j \le k$. So basic open subspaces of $\Omega _{\infty }$ are given by $\Omega _{\infty } \cap U(pP; p_1P, \dotsc , p_k P)$.

In general, two different infinite words can give rise to the same element in $\Omega _{\infty }$. However, this cannot happen for infinite words which contain no relator as a finite subword:

Lemma 2.4

For $w, \bar{w} \in \Sigma ^{\infty }$, let $w_i \mathrel {:=}w_{[1,i]}$ and $\bar{w}_j \mathrel {:=}\bar{w}_{[1,j]}$ for all $i, j \ge 1$. Then $\chi _w = \chi _{\bar{w}}$ if and only if for all i, there exists j with $\bar{w}_j \in w_i P$, and for all j, there exists i with $w_i \in \bar{w}_j P$.

Assume that the infinite word w does not contain any relator as a subword. Then $\chi _w = \chi _{\bar{w}}$ if and only if $w \equiv \bar{w}$.

Proof

The first part follows immediately from the construction of characters from infinite words. For the second part, $\chi _w = \chi _{\bar{w}}$ implies that given j, there exists i satisfying without loss of generality that $i \ge j$ such that $w_i = \bar{w}_j x$ for some $x \in \Sigma ^*$. Since $w_i$ contains no relator as a subword, we must have $w_i \equiv \bar{w}_j x$, and hence $w_j \equiv \bar{w}_j$. $\square$

3 The Non-reversible Case

Throughout this section, let P be a right LCM monoid given by a presentation $(\Sigma , R)$ as in Sect. 2.1.1 satisfying our standing assumptions in Sect. 2.1.2. Let G be the group given by the same presentation as P, and suppose that $P \rightarrow G$ induced by $\mathrm{id}_{\Sigma }$ is an embedding.

Let us start with the following general result which tells us that if our presentation does not have too many relations, then typically the boundary $\partial \Omega$ is as large as it can be.

Theorem 3.1

Suppose that $|\Sigma | \ge 2$. Assume that there is $\varepsilon \ne z \in \Sigma ^*$ such that z is not a subword of a relator, no relator is a subword of z, no prefix of z is a suffix of a relator, and no suffix of z is a prefix of a relator.

• (i) If there exist $a_1, \dotsc , a_n \in P$ such that $P {\setminus } \left\{ e \right\} = \bigcup _{i=1}^n a_iP$, then we have $\overline{G.\chi } = \Omega _{\infty }$ for every $\chi \in \Omega _{\infty }$.

• (ii) If we cannot write $P {\setminus } \left\{ e \right\} = \bigcup _{i=1}^n a_iP$ for some $a_1, \dotsc , a_n \in P$, then we have $\overline{G.\chi } = \Omega$ for every $\chi \in \Omega _{\infty }$.

Moreover, in either case there exists $\omega \in \Omega _{\infty }$ with trivial stabilizer group, $G_{\omega } = \left\{ e \right\}$.

Proof

Consider a basic open subset

$$\begin{aligned}U = U(pP; p_1 P, \dotsc , p_k P) = \left\{ \chi \in \Omega \text {: }\chi (pP) = 1, \, \chi (p_1 P) = \cdots = \chi (p_k P) = 0 \right\} ,\end{aligned}$$

with $p_j P \subsetneq pP$ for all $1 \le j \le k$. In case (i), we may assume that $\Omega _{\infty } \cap U \ne \emptyset$ and need to find $g \in G$ with $g.\chi \in U$ ($g.\chi$ will then automatically lie in $\Omega _{\infty } \cap U$ because $\chi \in \Omega _{\infty }$ and $\Omega _{\infty }$ is G-invariant). In case (ii), we only know that $U \ne \emptyset$, and we need to find $g \in G$ with $g.\chi \in U$.

Let us first of all show that in case (ii), we actually also have $\Omega _{\infty } \cap U \ne \emptyset$. Choose generators $\sigma _j \in \Sigma$ such that $p_j P \subseteq p \sigma _j P$. As $P {\setminus } \left\{ e \right\} \supsetneq \bigcup _j \sigma _jP$, there exists $\sigma \in \Sigma$ with $\sigma \notin \bigcup _j \sigma _j P$. Consider the infinite word $w = p \sigma z z z \cdots$, and the corresponding element $\chi _w$. Obviously, $\chi _w(pP) = 1$. If $\chi _w(p_j P) = 1$, then $p \sigma z^m \in p_j P$ for some m, which implies that $\sigma z^m \in \sigma _j P$. So there exists $x \in \Sigma ^*$ with $\sigma z^m = \sigma _j x$. Thus there exists a sequence $w_1, \dotsc , w_s \in \Sigma ^*$ such that $w_1 \equiv \sigma z^m$, $w_s \equiv \sigma _j x$, and for all $1 \le i \le s-1$, we have $w_i \equiv _R w_{i+1}$. By our assumptions on z, we must be able to write $w_r \equiv v_r z^m$ with $\ell ^*(v_r) \ge 1$ and $\sigma = v_r$ for all $1 \le r \le s$. Hence $v_s z^m \equiv \sigma _j x$, which implies that the first letter of $v_s$ must be $\sigma _j$, so that $v_s \in \sigma _j P$. However, we know that $\sigma = v_s$. This is a contradiction as $\sigma \notin \bigcup _j \sigma _j P$ by our choice of $\sigma$.

Now choose $\check{p}_j \in \Sigma ^*$ with $\check{p}_j = p_j$ in P. Let $\check{\Sigma } \subseteq \Sigma$ be a finite subset such that $\check{p}_j \in \check{\Sigma }^*$ for all j. Define $\check{\ell }^*$ to be the word length in $\check{\Sigma }^*$, and for $x \in P$, set $\check{\ell }(x) \mathrel {:=}\inf \left\{ L \in \mathbb {Z}_{\ge 0} \text {: }x = \sigma _1 \cdots \sigma _L, \, \sigma _1, \dotsc , \sigma _L \in \check{\Sigma } \right\}$. Note that if x cannot be represented by a word in $\check{\Sigma }^*$, then $\check{\ell }(x) = \infty$. We claim that in both cases (i) and (ii), we can always find $x \in pP$ with $x \notin p_j P$ for all j and $\check{\ell }(x) \ge \check{\ell }^*(\check{p}_j) = \ell ^*(\check{p}_j)$ for all j. In case (ii), this follows from our argument above and because $\lim _{m \rightarrow \infty } \ell (p \sigma z^m) = \infty$. In case (i), we can find $t \in \Sigma ^{\infty }$ with $\chi _t \in U$ because $\Omega _{\infty } \cap U \ne \emptyset$. Let $t_i \mathrel {:=}t_{[1,i]}$ be the word consisting of the first i letters of t. Then $t_1 P \supsetneq t_2 P \supsetneq t_3 P \supsetneq \cdots$, which implies that $\sup _i \check{\ell }(t_i) = \infty$. Otherwise $\left\{ t_i \text {: }i = 1, 2, \dotsc \right\}$ would be a finite set since $\check{\Sigma }$ is finite, but then we could not have $t_1 P \supsetneq t_2 P \supsetneq t_3 P \supsetneq \cdots$. Now choose i with $t_i \in pP$ and $\check{\ell }(t_i) \ge \max _j \check{\ell }^*(\check{p}_j)$, and set $x \mathrel {:=}t_i$.

So let $x \in pP$ satisfy $x \notin p_j P$ for all j and $\check{\ell }(x) \ge \check{\ell }^*(\check{p}_j)$ for all j. Set $g = xz$. We have to show that $g.\chi \in U$. Take $s \in \Sigma ^{\infty }$ with $\chi = \chi _s$, and let $s_i \mathrel {:=}s_{[1,i]}$ be the word consisting of the first i letters of s. Then $g.\chi = g.\chi _s = \chi _{xzs}$. Since $x \in pP$, we have $(g.\chi )(pP) = 1$. Assume that $(g.\chi )(p_j P) = 1$ for some j. Then there exists n with $x z s_n = \check{p}_j r$ for some $r \in \Sigma ^*$. This means that we can find $y_1, \dotsc , y_m \in \Sigma ^*$ with $y_1 \equiv x z s_n$, $y_m \equiv \check{p}_j r$ and for all $1 \le l \le m-1$, there exist $\alpha , u, v, \zeta \in \Sigma ^*$ with $(u,v) \in R$ or $(v,u) \in R$ or $(u,v) = (\varepsilon , \varepsilon )$ such that $y_l \equiv \alpha u \zeta$ and $y_{l+1} \equiv \alpha v \zeta$. By our assumptions on z, the subword z of $x z s_n$ will not be changed. Hence for every j, we must have $y_l \equiv x_l z s_{n,l}$, with $x_l = x$ and $s_{n,l} = s_n$ in P. In particular, we have $x_m z s_{n,m} \equiv \check{p}_j r$. Now we claim that $\ell ^*(x_m) \ge \ell ^*(\check{p}_j)$. If not, then $\ell ^*(x_m) < \ell ^*(\check{p}_j)$, so that $x_m \in \check{\Sigma }^*$, and thus $\ell ^*(x_m) = \check{\ell }^*(x_m) \ge \check{\ell }(x) \ge \check{\ell }^*(\check{p}_j) = \ell ^*(\check{p}_j)$, which is a contradiction. So we have $\ell ^*(x_m) \ge \ell ^*(\check{p}_j)$, which implies $x_m \equiv \check{p}_j q$ for some $q \in \Sigma ^*$. Hence it follows that $x \in \check{p}_j P = p_j P$, which is a contradiction. This shows $g.\chi \in U$.

Now let us find $\omega \in \Omega _{\infty }$ with trivial stabilizer group. First assume that z contains two different letters $\alpha$ and $\beta$. By assumption, $\alpha$ and $\beta$ are not relators. Choose a sequence $\sigma _1, \sigma _2, \dotsc$ in $\left\{ \alpha ,\beta \right\}$ such that the infinite word $\sigma _1 \sigma _2 \cdots$ has no period. Now form the infinite word $\bar{w} \mathrel {:=}z \sigma _1 z \sigma _2 z \cdots$. It has no period, either, because we cannot shift by a multiple of $\ell ^*(z) + 1$ as $\sigma _1 \sigma _2 \cdots$ has no period, and if shifting $\bar{w}$ by a different number of letters produces $\bar{w}$ again, then all the $\sigma _i$ would have to coincide with a fixed letter of z, which again is not possible as $\sigma _1 \sigma _2 \cdots$ has no period. By our assumption on z, $\bar{w}$ does not contain a relator as a finite subword. Now assume that $h \in G$ satisfies $h.\chi _{\bar{w}} = \chi _{\bar{w}}$. Then we must have $h = pq^{-1}$, and $\chi _{\bar{w}}(qP) = 1$. The latter implies that $z \sigma _1 z \sigma _2 \cdots z \sigma _n \in qP$ for some n, so that $z \sigma _1 z \sigma _2 \cdots z \sigma _n = qr$ in P. But since $z \sigma _1 z \sigma _2 \cdots z \sigma _n$ contains no relator as a subword, we must have $z \sigma _1 z \sigma _2 \cdots z \sigma _n \equiv qr$. In particular, $\bar{w} \equiv q \tilde{w}$ for some $\tilde{w} \in \Sigma ^{\infty }$, and $\tilde{w}$ also contains no relator as a finite subword and has no period. Now we have $\chi _{\bar{w}} = h.\chi _{\bar{w}} = h.\chi _{q \tilde{w}} = \chi _{p \tilde{w}}$. Since $\bar{w}$ contains no relator as a finite subword, Lemma 2.4 implies that $q \tilde{w} \equiv \bar{w} \equiv p \tilde{w}$. Since $\tilde{w}$ has no period, we must have $\ell ^*(p) = \ell ^*(q)$, which implies $p \equiv q$. Hence $h = pq^{-1} = e$.

Now assume that z is a word in a single letter. Then, as we assume $|\Sigma | \ge 2$, there must be another letter $\sigma \in \Sigma$ which does not appear in z. Form the infinite word $\bar{w} \mathrel {:=}z \sigma z^2 \sigma z^3 \sigma \cdots$. Assume that $h \in G$ satisfies $h.\chi _{\bar{w}} = \chi _{\bar{w}}$. Then we must have $h = pq^{-1}$, and $\chi _{\bar{w}}(qP) = 1$. Then there must exist $n \in \mathbb {Z}_{\ge 1}$ such that $z \sigma z^2 \sigma \cdots z^n \sigma = qr$ for some $r \in P$. Our assumptions on z imply that $q = z \sigma z^2 \sigma \cdots \sigma \zeta$ for some subword $\zeta$ of a power of z. So $\bar{w} = q \tilde{w}$ with $\tilde{w} = \eta \sigma z^k \sigma z^{k+1} \sigma \cdots$, where $\eta$ is a subword of some power of z. Now write $p = x_1 a_1 \cdots x_m a_m$, where $x_i$ are subwords of powers of z and $a_i$ are words not containing the letter appearing in z. As we have seen above, q is of the same form, i.e., $q = y_1 b_1 \cdots y_n b_n$, where $y_j$ are subwords of powers of z and $b_j$ are words not containing the letter appearing in z. We have $p \tilde{w} = x_1 a_1 \cdots x_m a_m \eta \sigma z^k \sigma z^{k+1} \sigma \cdots$ and $q \tilde{w} = y_1 b_1 \cdots y_n b_n \eta \sigma z^k \sigma z^{k+1} \sigma \cdots$. Now our assumptions on z imply that $m=n$ and $x_1 = y_1$, $a_1 = b_1$, ..., $x_m = y_m$ and $a_m = b_m$, and thus $p = q$, i.e., $h = p q^{-1} = e$.

$\square$

The following is an immediate consequence:

Corollary 3.2

Let us keep the same assumptions as in Theorem 3.1. If we are in case (i), i.e., there exist elements $a_1, \dotsc , a_n$ in our semigroup P such that $P {\setminus } \left\{ e \right\} = \bigcup _{i=1}^n a_iP$, then we have $\partial \Omega = \Omega {\setminus } P$ and, if G is exact, a short exact sequence $0 \rightarrow \mathcal K(\ell ^2 P) \rightarrow C^*_{\lambda }(P) \rightarrow \partial C^*_{\lambda }(P) \rightarrow 0$.

If we are in case (ii), i.e., we cannot write $P {\setminus } \left\{ e \right\} = \bigcup _{i=1}^n a_iP$ for some $a_1, \dotsc , a_n \in P$, then we have $\partial \Omega = \Omega$ and $C^*_{\lambda }(P) = \partial C^*_{\lambda }(P)$.

In both cases $\partial C^*_{\lambda }(P)$ is a purely infinite simple C*-algebra (we do not need exactness of G for this).

Here we used [26, Theorem 22.9]. The last part of the corollary follows from Theorem 3.1 together with [18, Corollary 5.7.17].

Remark 3.3

Let us assume that our presentation $(\Sigma ,R)$ is r-homogeneous in the sense of [21, Definition 4.1], or equivalently (see [21, Proposition 4.3]), that it is r-Noetherian in the sense of [21, Definition 2.6]. Then there exist elements $a_1, \dotsc , a_n$ in our semigroup $P = \left\langle \Sigma \, \vert \, R \right\rangle ^+$ such that $P {\setminus } \left\{ e \right\} = \bigcup _{i=1}^n a_iP$ if and only if $|\Sigma | < \infty$. The “if”-part is clear. For the “only if”-part, assume $|\Sigma | = \infty$ and that $P {\setminus } \left\{ e \right\} = \bigcup _{i=1}^n a_iP$ for some $a_i \in P$. Then P does not coincide with the monoid $\left\langle a_1, \dotsc , a_n \right\rangle ^+$ generated by $a_1, \dotsc , a_n$ (see Sect. 2.1.2). Take $x \in P {\setminus } \left\langle a_1, \dotsc , a_n \right\rangle ^+$. Define $x_0 \mathrel {:=}x$, and given $x_k \in P {\setminus } \left\langle a_1, \dotsc , a_n \right\rangle ^+$, we have $x_k \in \bigcup _{i=1}^n a_i P$ so that we can find $1 \le i \le n$ and $x_{k+1} \in P$ such that $x_k = a_i x_{k+1}$. Since $x_k \in P {\setminus } \left\langle a_1, \dotsc , a_n \right\rangle ^+$, we must have $x_{k+1} \in P {\setminus } \left\langle a_1, \dotsc , a_n \right\rangle ^+$. So we obtain a sequence $(x_k)$ such that $Px_0 \subsetneq Px_1 \subsetneq Px_2 \subsetneq \cdots$. This contradicts our assumption that $(\Sigma ,R)$ is r-Noetherian.

Let us now apply our findings to the one-relator case. From now on until the rest of this section, let $P = \left\langle \Sigma \, \vert \, u = v \right\rangle ^+$ be a right LCM one-relator monoid, with $(\Sigma , u = v)$ satisfying our standing assumptions in Sect. 2.1.2 and the assumptions in Sect. 2.1.3.

Lemma 3.4

Let $P = \left\langle \Sigma \, \vert \, u = v \right\rangle ^+$ with $|\Sigma | \ge 3$. Then there exists $\varepsilon \ne z \in \Sigma ^*$ such that z is not a subword of a relator, no relator is a subword of z, no prefix of z is a suffix of a relator, and no suffix of z is a prefix of a relator.

Proof

Let $c \in \Sigma$ be a generator which is not the first letter of u or v, and let $a \in \Sigma$ be a generator which is not the last letter of u or v. Consider $a^i c^j$ for $i, j > \max (\ell ^*(u), \ell ^*(v))$. Then no prefix of $a^i c^j$ is a suffix of a relator, and no suffix of $a^i c^j$ is a prefix of a relator. Moreover, $a^i c^j$ is not a subword of a relator. The only problem is that $a^i c^j$ could contain a relator as a subword. Certainly, $a^i c^j$ can contain at most one relator as a subword, say u. Then $u \equiv a^k c^l$ with $k, l \ge 1$. By our assumptions, this means that v does not start with a and does not end with c. Choose $b \in \Sigma$ with $b \not \equiv a$, $b \not \equiv c$. Then $a^i b c^j$ has all the desired properties. $\square$

Corollary 3.5

Let $P = \left\langle \Sigma \, \vert \, u = v \right\rangle ^+$ with $|\Sigma | \ge 3$. Then if $|\Sigma | < \infty$, we have $\partial \Omega = \Omega {\setminus } P$ and a short exact sequence $0 \rightarrow \mathcal K(\ell ^2 P) \rightarrow C^*_{\lambda }(P) \rightarrow \partial C^*_{\lambda }(P) \rightarrow 0$. If $|\Sigma | = \infty$, we have $\partial \Omega = \Omega$ and $C^*_{\lambda }(P) = \partial C^*_{\lambda }(P)$. In both cases, $\partial C^*_{\lambda }(P)$ is a purely infinite simple C*-algebra.

Proof

This follows from Lemma 3.4 and Corollary 3.2 once we show that for $P = \left\langle \Sigma \, \vert \, u = v \right\rangle ^+$, there exist elements $a_1, \dotsc , a_n \in P$ such that $P {\setminus } \left\{ e \right\} = \bigcup _{i=1}^n a_iP$ if and only if $|\Sigma | < \infty$. The “if”-part is clear. For the “only if”-part, assume $|\Sigma | = \infty$. Choose $\check{a}_i \in \Sigma ^*$ with $a_i = \check{a}_i$, let $\Sigma _A$ be the set of letters appearing in one of the $\check{a}_i$, and let $\Sigma _R$ be the set of letters which appear in u or v. Take $\tau \in \Sigma {\setminus } (\Sigma _A \cup \Sigma _R)$. Then $\tau \ne e$ and $\tau \notin \bigcup _{i=1}^n a_iP$ since otherwise, $\tau = \check{a}_i x$ (for some i and some $x \in P$) would imply that we can find a sequence $w_1, \dotsc , w_s$ in $\Sigma ^*$ with $w_1 \equiv \tau$ and $w_s \equiv \check{a}_i x$ such that $w_r \equiv _R w_{r+1}$ for all $1 \le r \le s-1$. But $\tau \equiv w_1 \equiv _R w_2$ implies $w_2 \equiv \tau$, and we conclude inductively that $w_s \equiv \tau$, hence $\check{a}_i x \equiv \tau$. This, however, is impossible. Hence $P {\setminus } \left\{ e \right\} \ne \bigcup _{i=1}^n a_iP$.

Note that exactness of the one-relator group $\left\langle \Sigma \, \vert \, u = v \right\rangle$ follows from [28]. $\square$

Remark 3.6

Let $P = \left\langle \Sigma \, \vert \, u = v \right\rangle ^+$ with $|\Sigma | \ge 3$, and let $G = \left\langle \Sigma \, \vert \, u = v \right\rangle$ be the group given by the same presentation as P. $P \subseteq G$ is Toeplitz in the sense of [18, § 5.8] if and only if $P \subseteq G$ is quasi-lattice ordered in Nica’s sense [41], i.e., for every $g \in G$, $P \cap gP = \emptyset$ or there exists $p \in P$ with $P \cap gP = pP$.

If $P \subseteq G$ is Toeplitz, then it follows from [17, Theorem 5.2] that $K_*(C^*_{\lambda }(P)) \cong K_*(\mathbb {C})$, and more precisely,

$$\begin{aligned}K_*(C^*_{\lambda }(P)) = (\mathbb {Z}[1],\left\{ 0 \right\} ).\end{aligned}$$

Here we use that one-relator groups satisfy the Baum–Connes conjecture with coefficients [3]. If in addition $3 \le |\Sigma | < \infty$, the short exact sequence $0 \rightarrow \mathcal K(\ell ^2 P) \rightarrow C^*_{\lambda }(P) \rightarrow \partial C^*_{\lambda }(P) \rightarrow 0$ induces in K-theory the exact sequence

$$\begin{aligned} 0 \rightarrow K_1(\partial C^*_{\lambda }(P)) \rightarrow \mathbb {Z}\rightarrow \mathbb {Z}\rightarrow K_0(\partial C^*_{\lambda }(P)) \rightarrow 0, \end{aligned}$$

where the homomorphism $\mathbb {Z}\rightarrow \mathbb {Z}$ is given by multiplication with $1 - |\Sigma | + 1 = 2 - |\Sigma |$ since the minimal projection $1_{\left\{ e \right\} } \in \mathcal K(\ell ^2 P)$ is given by $1_{\left\{ e \right\} } = 1_{P} - \sum _{\sigma \in \Sigma } 1_{\sigma P} + 1_{uP}$, and all the projections $1_{P}$, $1_{\sigma P}$ and $1_{uP}$ are Murray-von Neumann equivalent to 1 in $C^*_{\lambda }(P)$. Hence

$$\begin{aligned}(K_0(\partial C^*_{\lambda }(P)), [1], K_1(\partial C^*_{\lambda }(P))) \cong (\mathbb {Z}/ (|\Sigma | - 2) \mathbb {Z}, 1, \left\{ 0 \right\} ).\end{aligned}$$

If $|\Sigma | = \infty$, we have $C^*_{\lambda }(P) = \partial C^*_{\lambda }(P)$, so that

$$\begin{aligned}(K_0(\partial C^*_{\lambda }(P)), [1], K_1(\partial C^*_{\lambda }(P))) \cong (\mathbb {Z}, 1, \left\{ 0 \right\} ).\end{aligned}$$

Let us discuss another natural topic, namely nuclearity of $\partial C^*_{\lambda }(P)$, where $P = \left\langle \Sigma \, \vert \, u = v \right\rangle ^+$, with $|\Sigma | \ge 3$. Note that there are examples of such monoids for which the boundary quotient is not nuclear. Take for instance $P = \left\langle a, b, c \, \vert \, aba = bab \right\rangle ^+$. Then $P = B_3^+ * \left\langle c \right\rangle ^+$, where $B_3^+$ is the Braid monoid $B_3^+ = \left\langle a, b \, \vert \, aba = bab \right\rangle ^+$ and $\left\langle c \right\rangle ^+$ is the free monoid on one generator c. The Braid group $B_3 = \left\langle a, b \, \vert \, aba = bab \right\rangle$ appears as a stabilizer group for the partial action $G \curvearrowright \partial \Omega$, where $G = \left\langle a, b, c \, \vert \, aba = bab \right\rangle$. Hence $G \ltimes \partial \Omega$ is not amenable, and $\partial C^*_{\lambda }(P)$ is not nuclear.

Let us now present conditions on the presentation $(\Sigma , u=v)$ which ensure that $\partial C^*_{\lambda }(P)$ is nuclear. For this purpose, we introduce the following

Definition 3.7

Let $X \subseteq \Omega$ be a closed subset. The partial action $G \curvearrowright \Omega$ restricts to a partial dynamical system $G \curvearrowright X$, so that we can form the transformation groupoid $G \ltimes X$. Let $\mathcal E$ be a graph with vertices $\mathcal E^0$ and edges $\mathcal E^1$. Let $\varvec{\sigma }: \, \mathcal E^1 \rightarrow \Sigma$ be a map. $\varvec{\sigma }$ extends to maps on the finite path space $\mathcal E^* \rightarrow \Sigma ^*, \, \mu _1 \cdots \mu _n \mapsto \varvec{\sigma }(\mu _1) \cdots \varvec{\sigma }(\mu _n)$ and the infinite path space $\mathcal E^{\infty } \rightarrow \Sigma ^{\infty }, \, \mu _1 \mu _2 \cdots \mapsto \varvec{\sigma }(\mu _1) \varvec{\sigma }(\mu _2) \cdots$, both of which are again denoted by $\varvec{\sigma }$. Consider the composition $\varphi : \, \partial \mathcal E\rightarrow \Sigma ^* \cup \Sigma ^{\infty } \rightarrow \Omega$, where the first map is the restriction of $\varvec{\sigma }$ to $\partial \mathcal E\subseteq \mathcal E^* \cup \mathcal E^{\infty }$, and the second map is given by $\Sigma ^* \cup \Sigma ^{\infty } \rightarrow \Omega , \, w \mapsto \chi _w$. $(\mathcal E,\varvec{\sigma })$ is called a graph model for $G \curvearrowright X$ if $\Phi : \, \mathcal G_{\mathcal E} \rightarrow G \ltimes X, \, (\lambda \zeta , l(\lambda ) - l(\zeta ), \mu \zeta ) \mapsto (\varvec{\sigma }(\lambda ) \varvec{\sigma }(\mu )^{-1}, \varphi (\mu \zeta ))$ is an isomorphism of topological groupoids. Here $l(\cdot )$ denotes the path length in $\mathcal E^*$.

We refer the reader to [34, § 3.2] and the references therein for more information on how $\partial \mathcal E$ and $\mathcal G_{\mathcal E}$ are constructed. Note that we use the same convention as in [34, 45], which is different from the one in [48]. Moreover, the character $\chi _w$ attached to a word w has been constructed in Sects. 2.2.2 and 2.2.3.

Remark 3.8

Once $(\mathcal E,\varvec{\sigma })$ has been constructed, an important step in showing that $(\mathcal E,\varvec{\sigma })$ is a graph model is to show that $\varphi$ is a homeomorphism onto X. Note that, however, we cannot expect $\varvec{\sigma }$ to be injective on the set $\mathcal E^*$ of finite paths. The reason is that $\varvec{\sigma }$ will not remember the source of our finite paths. This problem will disappear once we go over to $\partial \mathcal E$, as $\partial \mathcal E$ only contains those finite paths in $\mathcal E^*$ whose source is a vertex z with $|r^{-1}(z)| = \infty$ or 0 (where $r(\cdot )$ is the range map of $\mathcal E$). The point is that in our graph models, there will be at most one vertex z with $|r^{-1}(z)| = \infty$, and for paths in $\mathcal E^{\infty }$, we can reconstruct sources by looking at subsequent parts of our infinite path.

Now let $P = \left\langle \Sigma \, \vert \, u = v \right\rangle ^+$ and $G = \left\langle \Sigma \, \vert \, u = v \right\rangle$. Assume that

$$\begin{aligned}&\mathrm{OVL}(v) = \left\{ x \in \Sigma ^* \text {: }v \equiv xy \equiv wx \ \mathrm{for} \ \mathrm{some} \ \varepsilon \ne w, y \in \Sigma ^* \right\} = \left\{ \varepsilon \right\} ,\end{aligned}$$

(1)

$$\begin{aligned}&\ell _{\sigma }(u) < \ell _{\sigma }(v) \ \mathrm{for} \ \mathrm{some} \ \sigma \in \Sigma ,\end{aligned}$$

(2)

$$\begin{aligned}&u \text { is not a subword of } v, \text { no prefix of } u \text { is a suffix of } v, \text { and no suffix of } u\nonumber \\&\quad \text { is a prefix of } v. \end{aligned}$$

(3)

We now construct graph models for $G \curvearrowright \partial \Omega$.

Theorem 3.9

Assume that (1), (2) and (3) are valid. Set $L \mathrel {:=}\ell ^*(v) - 1$.

• (i) Suppose that $3 \le |\Sigma | < \infty$. Define a graph as follows: Set $\mathcal E^0 \mathrel {:=}\left\{ x \in \Sigma ^* \text {: }\ell ^*(x) {=}L \right\}$ as the set of vertices, and let $\mathcal E^1 \mathrel {:=}\left\{ (y,x) \text {: }y, x \in \mathcal E^0; \; \exists \, \sigma , \tau \in \Sigma : \, y \tau \equiv \sigma x \not \equiv v \right\}$. Define range and source maps by $r(y,x) = y$ and $s(y,x) = x$. Moreover, define $\varvec{\sigma }: \, \mathcal E^1 \rightarrow \Sigma$ by setting $\varvec{\sigma }(y,x)$ as the first letter of y. Then $(\mathcal E,\varvec{\sigma })$ is a graph model for $G \curvearrowright \partial \Omega$.

• (ii) Suppose that $|\Sigma | = \infty$. Define a graph as follows: Let $\Sigma _R$ be the finite subset of $\Sigma$ containing all the letters that appear in the relators u and v. Set $\mathcal E^0 \mathrel {:=}\left\{ x \in \Sigma _R^* \text {: }\ell ^*(x) \le L \right\}$ as the set of vertices, and let

  $$\begin{aligned} \mathcal E^1 \mathrel {:=}&\left\{ (y,x) \text {: }y, x \in \mathcal E^0; \; \ell ^*(y) = \ell ^*(x) = L; \; \exists \, \sigma , \tau \in \Sigma _R: \, y \tau \equiv \sigma x \not \equiv v \right\} \\ \cup&\left\{ (y,x) \text {: }y, x \in \mathcal E^0; \; \ell ^*(y) = \ell ^*(x) + 1; \; \exists \, \sigma \in \Sigma _R: \, y \equiv \sigma x \right\} \\ \cup&\left\{ \mu _{\tau ,x} \text {: }x \in \mathcal E^0, \, \tau \in \Sigma {\setminus } \Sigma _R \right\} . \end{aligned}$$

  Define range and source maps by $r(y,x) = y$, $s(y,x) = x$, and $r(\mu _{\tau ,x}) = \varepsilon$, $s(\mu _{\tau ,x}) = x$. Moreover, define $\varvec{\sigma }: \, \mathcal E^1 \rightarrow \Sigma$ by setting $\varvec{\sigma }(y,x)$ as the first letter of y, and $\varvec{\sigma }(\mu _{\tau ,x}) = \tau$. Then $(\mathcal E,\varvec{\sigma })$ is a graph model for $G \curvearrowright \Omega$.

Note that $3 \le |\Sigma |$ ensures that our monoid is not left reversible (see Sect. 2.1.5).

Proof

(i): First of all, Corollary 3.5 implies that in this case, we have $\partial \Omega = \Omega _{\infty }$. As explained in Sect. 2.1.5, our conditions (1) and (2) imply that for every $z \in \Sigma ^*$, we can find a unique irreducible word $x \in \Sigma ^*$ (i.e., x does not contain v as a subword) such that $z = x$ in P, and two irreducible words $x, y \in \Sigma ^*$ represent the same element in P if and only if $x \equiv y$. Given $z \in \Sigma ^*$, we write $\rho (z)$ for the unique irreducible word in $\Sigma ^*$ with $\rho (z) = z$ in P. Given $\chi \in \Omega _{\infty }$, Lemma 2.3 implies that we can find $x \in \Sigma ^{\infty }$ such that $\chi = \chi _x$. Then x can be chosen to be of the form $x \equiv y_1 v y_2 v y_3 v \cdots$, where $y_i \in \Sigma ^*$ do not contain v as a subword. Set $w \mathrel {:=}y_1 u y_2 u y_3 u \cdots$. Then w does not contain v as a finite subword by our assumptions on u and v, and we have $\chi = \chi _x = \chi _w$ by Lemma 2.4 (since $x_i = w_i$ for infinitely many i, with $w_i = w_{[1,i]}$ being the word consisting of the first i letters of w). Moreover, given infinite words $\tilde{w}, \bar{w} \in \Sigma ^{\infty }$ which do not contain v as finite subword such that $\chi _{\tilde{w}} = \chi _{\bar{w}}$, then Lemma 2.4 implies that for all j, there exists i such that $\tilde{w}_i \in \bar{w}_j P$, where $\tilde{w}_i = \tilde{w}_{[1,i]}$ and $\bar{w}_j = \bar{w}_{[1,j]}$ (the words consisting of the first i or j letters of $\tilde{w}$ or $\bar{w}$, respectively). In particular, for $j \ge l \mathrel {:=}\ell ^*(v)$, because of our assumption on u and v, we conclude from $\tilde{w}_i \in \bar{w}_j P$ that $i \ge j-l$ and $\tilde{w}_{j-l} \equiv \bar{w}_{j-l}$. Hence $\tilde{w} \equiv \bar{w}$. This shows that we have a bijection

$$\begin{aligned} \left\{ w \in \Sigma ^{\infty } \text {: }w \text { is } v \text {-free} \right\} \cong \Omega _{\infty }, \, w \mapsto \chi _w. \end{aligned}$$

Here $w \in \Sigma ^{\infty }$ is called v-free if w does not contain v as finite subword. To see that $\varvec{\sigma }$ gives rise to a bijection $\mathcal E^{\infty } \cong \left\{ w \in \Sigma ^{\infty } \text {: }w \text { is } v \text {-free} \right\}$, we construct an inverse as follows: Given a v-free $w \in \Sigma ^{\infty }$, say $w {=} \sigma _1 \sigma _2 \sigma _3 \cdots$ with $\sigma _i \in \Sigma$, let $x_i \mathrel {:=}\sigma _i \cdots \sigma _{i+L-1}$, set $\mu _i \mathrel {:=}(x_i,x_{i+1})$, and define $\varvec{\mu }(w) \mathrel {:=}\mu _1 \mu _2 \mu _3 \cdots$. In this way, we obtain a map $\varvec{\mu }: \, \left\{ w \in \Sigma ^{\infty } \text {: }w \text { is } v \text {-free} \right\} \rightarrow \mathcal E^{\infty }$ which is the inverse of $\mathcal E^{\infty } \rightarrow \left\{ w \in \Sigma ^{\infty } \text {: }w \text { is } v \text {-free} \right\} , \mu \mapsto \varvec{\sigma }(\mu )$. Therefore, $\varphi$ is bijective as it is a composition of two bijections. Moreover, it is easy to see that $\varphi$ is a homeomorphism.

Clearly, $\Phi$ is a continuous groupoid homomorphism. To construct its inverse, define $\Psi : \, G \ltimes \partial \Omega \rightarrow \mathcal G_{\mathcal E}, \, (g,\chi _w) \mapsto (\varphi ^{-1}(g.\chi _w), \lim _{n \rightarrow \infty } (\ell ^*(\rho (g w_n)) {-} \ell ^*(w_n)), \varphi ^{{-}1}(\chi _w))$, where w is a v-free word, and $w_n = w_{[1,n]}$ (the word consisting of the first n letters of w). To show that $\Psi$ is a well-defined continuous groupoid homomorphism, let us show that $\ell ^*(\rho (g w_n)) - \ell ^*(w_n)$ is well-defined and eventually constant (i.e., it does not depend on n as long as n is big enough). First of all, $\chi _w \in U_{g^{-1}}$ implies that there exists $M \in \mathbb {Z}_{\ge 0}$ such that $g w_M \in P$. Moreover, our assumptions imply that there exists an integer $N \ge M$ ($N = M + L$ works) such that for all integers $n \ge N$, we have $\rho (g w_n) = \rho (g w_N) w_{[N+1,n]}$, where $w_{[N+1,n]}$ is the subword of w starting with the $(N+1)$-th letter and ending with the n-th letter. Thus we have for all $n \ge N$ that $\ell ^*(\rho (g w_n)) - \ell ^*(w_n) = \ell ^*(\rho (g w_N) w_{[N+1,n]}) - \ell ^*(w_N w_{[N+1,n]}) = \ell ^*(\rho (g w_N)) - \ell ^*(w_N)$. So, indeed, it does not depend on n. This shows that $\Psi$ is a well-defined continuous groupoid homomorphism. It is straightforward to see that $\Psi$ is the inverse of $\Phi$.

(ii): We have $\Omega = P \cup \Omega _{\infty }$. As in (i), we have a bijection

$$\begin{aligned} \left\{ w \in \Sigma ^* \cup \Sigma ^{\infty } \text {: }w \text { is } v \text {-free} \right\} \cong \Omega , \, w \mapsto \chi _w. \end{aligned}$$

We claim that $\varvec{\sigma }$ gives rise to a bijection $\partial \mathcal E\cong \left\{ w \in \Sigma ^* \cup \Sigma ^{\infty } \text {: }w \text { is } v \text {-free} \right\}$. First of all, note that $\partial \mathcal E$ consists of the finite paths in $\mathcal E$ which start in $\varepsilon$ (i.e., which have source equal to $\varepsilon$) and the infinite paths in $\mathcal E$. We now set out to construct an inverse of $\varvec{\sigma }$. Given a v-free $w \in \Sigma _R^* \cup \Sigma _R^{\infty }$, say $w = \sigma _1 \sigma _2 \sigma _3 \cdots$ with $\sigma _i \in \Sigma _R$, let $x_i \mathrel {:=}\sigma _i \cdots \sigma _{\max (i+L-1, \, l(w))}$, set $\mu _i \mathrel {:=}(x_i,x_{i+1})$, and define $\varvec{\mu }(w) \mathrel {:=}\mu _1 \mu _2 \mu _3 \cdots$ ($\varvec{\mu }(w) \mathrel {:=}\mu _1 \mu _2 \mu _3 \cdots \mu _{l(w)}$ if $\ell ^*(w) < \infty$). A general v-free word $w \in \Sigma ^* \cup \Sigma ^{\infty }$ is of the form $w = w_1 \tau _1 w_2 \tau _2 \cdots$ or $w = w_1 \tau _1 w_2 \tau _2 \cdots z$, where $w_i$ are v-free words in $\Sigma _R^*$, $\tau _i \in \Sigma {\setminus } \Sigma _R$, and z is a v-free word in $\Sigma _R^{\infty }$. In the first case, set $\varvec{\mu }(w) = \varvec{\mu }(w_1) \mu _{\tau _1, \, r(\varvec{\mu }(w_2))} \varvec{\mu }(w_2) \mu _{\tau _2, \, r(\varvec{\mu }(w_3))} \cdots$, and in the second case, set $\varvec{\mu }(w) = \varvec{\mu }(w_1) \mu _{\tau _1, \, r(\varvec{\mu }(w_2))} \varvec{\mu }(w_2) \mu _{\tau _2, \, r(\varvec{\mu }(w_3))} \cdots \varvec{\mu }(z)$. In this way, we obtain a map $\varvec{\mu }: \, \left\{ w \in \Sigma ^* \cup \Sigma ^{\infty } \text {: }w \text { is } v \text {-free} \right\} \rightarrow \partial \mathcal E$ which is the inverse of $\partial \mathcal E\rightarrow \left\{ w \in \Sigma ^* \cup \Sigma ^{\infty } \text {: }w \text { is } v \text {-free} \right\} , \mu \mapsto \varvec{\sigma }(\mu )$. Therefore, $\varphi$ is bijective as it is a composition of two bijections. Moreover, it is easy to see that $\varphi$ is a homeomorphism.

Again, $\Phi$ is a continuous groupoid homomorphism, so we just need to construct its inverse $\Psi : \, G \ltimes \partial \Omega \rightarrow \mathcal G_{\mathcal E}$. For a v-free word $w \in \Sigma ^{\infty }$, we define $\Psi (g,\chi _w)$ as in (i). For a v-free word $w \in \Sigma ^*$, define $\Psi (g,\chi _w) \mathrel {:=}(\varphi ^{-1}(g.\chi _w), \ell ^*(\rho (gw)) - \ell ^*(w), \varphi ^{-1}(\chi _w))$. Then $\Psi$ is a well-defined continuous groupoid homomorphism which is the inverse of $\Phi$. $\square$

Combining Theorem 3.9 with Corollary 3.5, we obtain

Corollary 3.10

Let $P = \left\langle \Sigma \, \vert \, u = v \right\rangle ^+$ with $|\Sigma | \ge 3$. Assume that (1), (2) and (3) hold. Let $\mathcal E$ be the graph model constructed in Theorem 3.9. Then the boundary quotient of $C^*_{\lambda }(P)$ is isomorphic to the graph C*-algebra of $\mathcal E$, $\partial C^*_{\lambda }(P) \cong C^*(\mathcal E)$, and it is a UCT Kirchberg algebra. In particular, $C^*_{\lambda }(P)$ and $\partial C^*_{\lambda }(P)$ are nuclear, and $G \curvearrowright \Omega$ is amenable (as is $G \ltimes \Omega$).

Example 3.11

Here is a concrete class of examples where Corollary 3.10 applies: Let A and B be sets with $|A| + |B| \ge 3$. Choose $u \in A^*$ arbitrary and $v \in B^*$ with $\mathrm{OVL}(v) = \left\{ \varepsilon \right\}$. Then the presentation $(A \cup B, u = v)$ satisfies all the assumptions in Theorem 3.9 and Corollary 3.10, so that Corollary 3.10 applies to the monoid $P = \left\langle A \cup B \, \vert \, u = v \right\rangle ^+$.

4 The Reversible Case

Throughout this section, let P be left reversible. In that case, $\partial \Omega$ reduces to a point, given by the map $\mathcal J\rightarrow \left\{ 0,1 \right\}$ sending every non-empty set in $\mathcal J$ to 1. We denote this point by $\infty$. Let us also assume that P is right LCM throughout this section. Recall that $\Omega _{\infty } = \Omega {\setminus } P$. In the following, we write $\tilde{\Omega } \mathrel {:=}\Omega {\setminus } \left\{ \infty \right\}$ and $\tilde{\Omega }_{\infty } \mathrel {:=}\Omega _{\infty } {\setminus } \left\{ \infty \right\}$.

4.1 Reversible one-relator monoids

Let us present one-relator monoids $P = \left\langle \Sigma \, \vert \, u=v \right\rangle ^+$ which are left reversible and for which we can describe $G \ltimes \tilde{\Omega }_{\infty }$. Let $(\Sigma , u = v)$ satisfy the assumptions in Sects. 2.1.2 and 2.1.3.

As explained in Sect. 2.1.5, $P = \left\langle \Sigma \, \vert \, u=v \right\rangle ^+$ can only be left reversible if $|\Sigma | \le 2$. Since our assumptions force $|\Sigma | \ge 2$, we must have $|\Sigma | = 2$, say $\Sigma = \left\{ a,b \right\}$. The next lemma gives a special condition which enforces left reversibility.

Lemma 4.1

If there is $w \in P$ with $w = a x = x x' = b y = y y'$ for some $x, y, x', y' \in P$, then we have

• $pw \in wP$ for all $p \in P$;

• for every $p \in P$, there exists $i \ge 0$ with $w^i \in pP$.

Proof

To prove the first claim, take $p \in \left\{ a,b \right\} ^*$. We proceed inductively on $\ell ^*(p)$. Our claim certainly holds for $p = \varepsilon$. If $pw \in wP$, then $paw = p a x x' = pw x' \in wP$. Similarly, $pbw \in wP$.

To prove the second claim, again take $p \in \left\{ a,b \right\} ^*$ and proceed inductively on $\ell ^*(p)$. Our claim certainly holds for $p = \varepsilon$. Assume $w^i \in pP$, say $w^i = pq$ for some $q \in P$. By (i), we know that $qw = wr$ for some $r \in P$. Hence $p a x r = p w r = p q w = w^{i+1}$. This shows that $w^{i+1} \in paP$. Similarly, $w^{i+1} \in pbP$. $\square$

For instance, if P is a Garside monoid in the sense of [20, 22, 23], then we may take for w a Garside element in P.

The following is now an immediate consequence of Lemma 4.1.

Corollary 4.2

If there is $w \in P$ with $w = a \alpha = \alpha \gamma = b \beta = \beta \delta$ for some $\alpha , \beta , \gamma , \delta \in P$, then $\chi \in \Omega$ satisfies $\chi (pP) = 1$ for all $p \in P$ if and only if $\chi (w^i P) = 1$ for all $i \ge 0$. Moreover, let $X_i \mathrel {:=}\left\{ \chi \in \Omega \text {: }\chi (w^i P) = 1, \, \chi (w^{i+1} P) = 0 \right\}$. Then $\Omega$ admits the following decomposition as a (set-theoretic) disjoint union

$$\begin{aligned} \Omega = \Big ( \coprod _{i=0}^{\infty } X_i \Big ) \amalg \left\{ \infty \right\} . \end{aligned}$$

We write $X \mathrel {:=}X_0$. Our goal is to describe $G \ltimes \tilde{\Omega }_{\infty }$. Since $Y \mathrel {:=}\Omega _{\infty } \cap X$ is obviously a compact open subspace of $\Omega _{\infty }$ which meets every G-orbit of $\tilde{\Omega }_{\infty }$, we have $G \ltimes \tilde{\Omega }_{\infty } \sim _M (G \ltimes \tilde{\Omega }_{\infty }) \, \vert \, Y$. Hence it suffices to describe the restriction $(G \ltimes \tilde{\Omega }_{\infty }) \, \vert \, Y$ of $G \ltimes \tilde{\Omega }_{\infty }$ to Y. Let us now—in the same spirit as Theorem 3.9—present two cases where we can provide a graph model for $G \curvearrowright Y$.

Theorem 4.3

• I. Assume that in Lemma 4.1, we may take $w = u = v$. Let $L \mathrel {:=}\max (\ell ^*(u), \ell ^*(v)) - 1$. Define a graph as follows: As $\mathcal E^0$, take the (finite) collection of all finite subwords of length L in u, v-free infinite words. Let $\mathcal E^1 \mathrel {:=}\left\{ (y,x) \text {: }y \right\} , x \in \mathcal E^0; \; \exists \, \sigma , \tau \in \left\{ a,b \right\} : \, y \tau \equiv \sigma x \text { is } u, v \text {-free}$. Define range and source maps by $r(y,x) = y$ and $s(y,x) = x$. Moreover, define $\varvec{\sigma }: \, \mathcal E^1 \rightarrow \left\{ a,b \right\}$ by setting $\varvec{\sigma }(y,x)$ as the first letter of y. Then $(\mathcal E,\varvec{\sigma })$ is a graph model for $G \curvearrowright Y$.

• II. Now we do not assume that w in Lemma 4.1 satisfies $w = u = v$. Suppose the following hold:

  • II.1. Let $\mathcal X\mathrel {:=}\left\{ z \in \left\{ a,b \right\} ^* \ v \text {-free} \text {: }z \in wP \right\}$. Assume that there exists a finite subset $\mathcal W\subseteq \mathcal X$ such that for all $z \in \mathcal X$ there is $x \in \mathcal W$ such that x is a subword of z.

  • II.2. u is not a subword of v, no prefix of u is a suffix of v, no suffix of u is a prefix of v, and v is not a subword of u.

  • II.3. There exists $k \ge 0$ such that for all $z = ar \in \left\{ a,b \right\} ^*$ and $\bar{z} = bs \in \left\{ a,b \right\} ^*$ with no v as a subword and with $\ell ^*(r), \ell ^*(s) \ge k$, we must have $zP \cap \bar{z}P \subseteq wP$.

  Let $L \mathrel {:=}\max (\left\{ \ell ^*(x) \text {: }x \in \left\{ v \right\} \cup \mathcal W \right\} ) - 1$. Define a graph as follows: As the vertex set $\mathcal E^0$, take the (finite) collection of all finite subwords of length L in $v, \mathcal W$-free infinite words. For the edges, set $\mathcal E^1 \mathrel {:=}\left\{ (y,x) \text {: }y \right\} , x \in \mathcal E^0; \; \exists \, \sigma , \tau \in \left\{ a,b \right\} : \, y \tau \equiv \sigma x \text { is } v, \mathcal W\text {-free}$. Define range and source maps by $r(y,x) = y$ and $s(y,x) = x$. Moreover, define $\varvec{\sigma }: \, \mathcal E^1 \rightarrow \left\{ a,b \right\}$ by setting $\varvec{\sigma }(y,x)$ as the first letter of y. Then $(\mathcal E,\varvec{\sigma })$ is a graph model for $G \curvearrowright Y$.

Here a word (finite or infinite) is called u, v-free if it does not contain u or v as a finite subword, v-free if it does not contain v as a finite subword, and $v, \mathcal W$-free if it is v-free and contains no $x \in \mathcal W$ as a finite subword.

Proof

In case I., note that we have a bijection $\left\{ z \in \left\{ a,b \right\} ^{\infty } \text {: }z \text { is } u,v \text {-free} \right\} \cong Y, \, z \mapsto \chi _z$ by Lemma 2.4. Moreover, the inverse for the map $\mathcal E^{\infty } \rightarrow \left\{ z \in \left\{ a,b \right\} ^{\infty } \text {: }z \right\} \text { is } u, v \text {-free}, \, \mu \mapsto \varvec{\sigma }(\mu )$ is given as follows: Given a u, v-free $z \in \left\{ a,b \right\} ^{\infty }$, say $z = \sigma _1 \sigma _2 \sigma _3 \cdots$ with $\sigma _i \in \left\{ a,b \right\}$, let $x_i \mathrel {:=}\sigma _i \cdots \sigma _{i+L-1}$, set $\mu _i \mathrel {:=}(x_i,x_{i+1})$, and define $\varvec{\mu }(w) \mathrel {:=}\mu _1 \mu _2 \mu _3 \cdots$. Hence the map $\varphi$ from Definition 3.7—being the composition of two bijections – is bijective. It is easy to see that $\varphi$ is a homeomorphism. Finally, the inverse of the map $\Phi$ from Definition 3.7 is given by $\Psi : \, G \ltimes Y \rightarrow \mathcal G_{\mathcal E}, \, (g,\chi _z) \mapsto (\varphi ^{-1}(g.\chi _z), \lim _{n \rightarrow \infty } (\ell ^*(g z_n) - \ell ^*(z_n)), \varphi ^{-1}(\chi _z))$, where z is a u, v-free word, and $z_n = z_{[1,n]}$ is the word consisting of the first n letters of z.

In case II., note that as in the proof of Theorem 3.9 (i), given an infinite word, condition II.2 guarantees that we can always replace v by u to arrive at a v-free infinite word.

Condition II.1 implies that for a v-free infinite word z, $\chi _z(wP) = 0$ if and only if z is $\mathcal W$-free: Suppose that z is not $\mathcal W$-free, i.e., it is of the form $z \equiv \alpha x \zeta$ with $x \in \mathcal W$. As x lies in wP, we can write $x = wy$ for some $y \in P$. Hence $\alpha x = \alpha w y \in wP$ by Lemma 4.1. By construction of $\chi _z$ (see Sect. 2.2.3), this implies that $\chi _z(wP) = 1$. Conversely, suppose that $\chi _z(wP) = 1$. Again by construction of $\chi _z$ (see Sect. 2.2.3), there must exist a finite word $z'$ with $z' \in wP$ such that $z \equiv z'z''$. Since z is v-free, so is $z'$. Hence $z'$ lies in $\mathcal X$, and condition II.1 implies that there exists $x \in \mathcal W$ such that x is a subword of $z'$. Hence x is a subword of z, so that z is not $\mathcal W$-free.

Condition II.3. ensures that two distinct v-free infinite words x and y determine distinguished characters $\chi _x \ne \chi _y \in Y$: Let x, y be v-free infinite words with $x \not \equiv y$, say $x \equiv p a \cdots$ and $y \equiv p b \cdots$ for some $p \in \left\{ a,b \right\} ^*$. For some i and j, we have $x_i \equiv x_{[1,i]} \equiv par$ and $y_j \equiv y_{[1,j]} \equiv pbs$ for some $r, s \in \left\{ a,b \right\} ^*$ with $\ell ^*(r), \ell ^*(s) \ge k$ ($x_{[1,i]}$ and $y_{[1,j]}$ are the words consisting of the first i and j letters of x and y, respectively). Now suppose that $\chi _x = \chi _y \mathrel {=:}\chi$, and hence that $\chi (x_i P) = 1 = \chi (y_j P)$. Then $\chi ((parP) \cap (pbsP)) = 1$. Condition II.3. implies that $(parP) \cap (pbsP) \subseteq pwP \subseteq wP$, where the last inclusion is due to Lemma 4.1 (i). Hence $\chi (wP) = 1$, so that $\chi \notin Y$.

Therefore, we have a bijection $\left\{ z \in \left\{ a,b \right\} ^{\infty } \text {: }z \text { is } v, \mathcal W\text {-free} \right\} \cong Y, \, z \mapsto \chi _z$. As in case I., we deduce that $\varphi$ from Definition 3.7 is a homeomorphism. The proof that the map $\Phi$ from Definition 3.7 is an isomorphism of topological groupoids is now similar to the one of Theorem 3.9 (i). $\square$

Remark 4.4

Once a graph model has been found, it will follow that $(G \ltimes \tilde{\Omega }_{\infty }) \, \vert \, Y$ and $G \ltimes \tilde{\Omega }_{\infty }$ are amenable, and thus $C^*_r((G \ltimes \tilde{\Omega }_{\infty }) \, \vert \, Y)$ and $C^*_r(G \ltimes \tilde{\Omega }_{\infty })$ are nuclear (see for instance [34, Remark 3.6]).

Example 4.5

Here are two classes of examples for case I: Consider $u = aba \cdots$ and $v = bab \cdots$ with $\ell ^*(u) = \ell ^*(v)$, or consider $u = a^p$ and $v = b^q$ for $p, q \ge 2$. The corresponding presentations $(\left\{ a,b \right\} , u=v)$ both define Garside monoids, where $w = u = v$ is a Garside element. Thus the conditions in case I of Theorem 4.3 are satisfied. The groups defined by these presentations are the Artin–Tits groups of dihedral type and the torus knot groups. It is conjectured in [43] that these are the only Garside groups on two generators (see also [22, Question 1]).

Remark 4.6

For all the presentations in Example 4.5 apart from $(\left\{ a,b \right\} , ab=ba)$ and $(\left\{ a,b \right\} , a^2 = b^2)$, $C^*_r((G \ltimes \tilde{\Omega }_{\infty }) \, \vert \, Y)$ and $C^*_r(G \ltimes \tilde{\Omega }_{\infty })$ are purely infinite simple. This is straightforward to check using, for instance, [48, Theorem 3.15]. (Note that the conventions in [34, 48] are different, one has to reverse the arrows in the graphs in one of these papers to translate into the convention of the other.) Therefore, $C^*_r((G \ltimes \tilde{\Omega }_{\infty }) \, \vert \, Y)$ and $C^*_r(G \ltimes \tilde{\Omega }_{\infty })$ are UCT Kirchberg algebras. While $C^*_r((G \ltimes \tilde{\Omega }_{\infty }) \, \vert \, Y)$ is unital, $C^*_r(G \ltimes \tilde{\Omega }_{\infty })$ is not unital and hence must be stable by Zhang [52]. As UCT Kirchberg algebras are classified by their K-theory, it is a natural task to compute K-theory for $C^*_r((G \ltimes \tilde{\Omega }_{\infty }) \, \vert \, Y)$ and $C^*_r(G \ltimes \tilde{\Omega }_{\infty })$. We will discuss K-theory in Sect. .

In order to discuss some examples for case II, let us formulate a stronger condition which implies condition II.3. Assume that u starts with b and v starts with a, i.e., $u = b \cdots$ and $v = a \cdots$. For $1 \le l \le \ell ^*(v) - 1$, let $v_{[1,l]}$ be the word consisting of the first l letters of v, and let $\sigma _{l+1} \in \left\{ a,b \right\}$ be the generator which is not the $(l+1)$-th letter of v. Now consider the following condition:

• II.3’. For all $1 \le l \le \ell ^*(v) - 1$, $v_{[1,l]} \sigma _{l+1} P \cap bP \subseteq wP$.

Clearly, condition II.3’. implies II.3. (with $k = \ell ^*(v)$).

Example 4.7

Let $u = b^j$ and $v = a b^i a b^i a \cdots a b^i a$, with $i > 0$ and $\ell _b(u) > \ell _b(v)$. Then we saw in Sect. 2.1.4 that $P = \left\langle a,b \, \vert \, u = v \right\rangle ^+$ is right LCM. It is easy to see that $w = b^{i+j}$ satisfies the conditions in Lemma 4.1. Moreover, our assumptions on u and v in case II are satisfied: Clearly, condition II.2 is satisfied.

It is easy to check that $W {\mathrel {:=}}\left\{ x {\in } \left\{ a,b \right\} ^* \text {: }x {=} w \ \mathrm{in} \ P \right\} {=} \left\{ b^{i+j} \right\} \cup \left\{ b^h v b^{i-h} \text {: }0 {\le } h {\le } i \right\}$. The only v-free word in W is given by $b^{i+j}$. Let us verify condition II.1 for $\mathcal W\mathrel {:=}\left\{ x \in \left\{ a,b \right\} ^* \ v \text {-free}x = w \ \mathrm{in} \ P \text {: }= \right\}$ $\left\{ b^{i+j} \right\}$. Take $z \in \mathcal X$. As explained in Sect. 2.1.1, $z \in wP$ means that there exist $z_1$, ..., $z_s$ in $\left\{ a,b \right\} ^*$ such that $z_1 \equiv z$, $z_{n+1}$ is obtained from $z_n$ by replacing (once) u by v or v by u, and $z_s$ contains some $y \in W$ as a subword. Without loss of generality we may assume that s is minimal. Let us show that z must already contain $b^{i+j}$ as a subword. Let $r \in \left\{ 1, \dotsc , s \right\}$ be maximal such that for all $1 \le m \le r-1$, $z_{m+1}$ is obtained from $z_m$ by replacing (once) u by v. Assume first that $r=s$. Condition II.2 implies that we must have $z \equiv \cdots u \cdots u \cdots u \cdots$, and $z_r \equiv z_s \equiv \cdots v \cdots v \cdots v \cdots$, i.e., $z_r \equiv z_s$ is obtained from z by replacing finitely many copies of u by v, and in each step, changes take place in a different part of z (there can be no overlap between a newly created v and the next u which is replaced because of condition II.2). Assume that $r > 1$. Since $z_r \equiv z_s$ contains $y \in W$ as a subword, and because by minimality of s we know that $z_{r-1}$ cannot contain some word in W as a subword, y must have an overlap with a newly created v, i.e., $z_r$ is of the form $\cdots b^h v b^{i-h} \cdots$ or $\cdots a b^i v \cdots$ or $\cdots v b^i a \cdots$, and $z_{r-1}$ is of the form $\cdots b^h u b^{i-h} \cdots$ or $\cdots a b^i u \cdots$ or $\cdots u b^i a \cdots$. In each of these cases, we see that $z_{r-1}$ already contains some word in W as a subword, contradicting minimality of s. Hence we conclude $r=1$, i.e., already z contains some word in W as a subword. As z is v-free, it follows that z must contain $b^{i+j}$ as a subword, as claimed. Now assume $r < s$. Then $z_{r+1}$ is obtained from $z_r$ by replacing (once) v by u. So $z_r$ contains v as a subword. We may as well assume that this instance of v was not created as a replacement $u \rightarrow v$ in going from $z_c$ to $z_{c+1}$ for some $c < r$. Since z is v-free, part of this copy of v must have been newly created when going from $z_q$ to $z_{q+1}$ by replacing (once) u by v, for some $1 \le q \le r-1$. So we must have $z_q \equiv \cdots u \zeta \cdots$ and $z_{q+1} \equiv \cdots v \zeta \cdots \equiv \cdots v' v'' \zeta \cdots$ with $v \equiv v'v''$ and $v'' \zeta \equiv v$, where $\zeta$ is a proper subword of v. (Another possibility would be $z_q \equiv \cdots \zeta u \cdots$, which is treated in an analogous way.) As $\zeta \not \equiv v$, $v''$ cannot be empty, and hence $v''$ must end with a since v ends with a. Because s is minimal, $\zeta$ cannot be empty. Therefore, $\zeta$ must start with $b^i$, and hence $b^{i+j}$ is a subword of $u \zeta$, which in turn is a subword of $z_q$, contradicting minimality of s. So all in all, we have proven that z must contain $b^{i+j}$ as a subword, as desired.

To verify condition II.3’., note that $v_{[1,l]} \sigma _{l+1}$ is either of the form $a b^i a b^i a \cdots a b^{i+1}$ (with less letters a appearing as in v) or $a b^i a b^i a \cdots a b^{h} a$ for $h < i$ (with at most as many letters a appearing as in v). In both cases, it is easy to check – for instance with the help of right reversing – that $v_{[1,l]} \sigma _{l+1} P \cap bP \subseteq b^{i+j} P$.

Remark 4.8

As in Remark 4.6, it is straightforward to check that for the presentations in Example 4.7, $C^*_r((G \ltimes \tilde{\Omega }_{\infty }) \, \vert \, Y)$ is a unital UCT Kirchberg algebra and $C^*_r(G \ltimes \tilde{\Omega }_{\infty })$ is a stable UCT Kirchberg algebra.

Remark 4.9

There is an overlap between the two classes of groups in Example 4.5, because the presentations $(\left\{ a,b \right\} , aba \cdots = bab \cdots )$, where each of the relators has m factors, with m odd, and $(\left\{ a,b \right\} , a^2=b^m)$ define isomorphic groups. Moreover, the groups given by the presentations in Example 4.7 are not new; they turn out to be isomorphic to the torus knot groups in Example 4.5.

4.2 Artin–Tits monoids of finite type

Let us now study Artin–Tits monoids of finite type with more than two generators (see [6, 24, 47]). We will see that the results in Sect. 4.1 do not carry over. Indeed, $G \ltimes \tilde{\Omega }_{\infty }$ is typically not amenable. Moreover, our structural results for $C^*_r(G \ltimes \tilde{\Omega }_{\infty })$ and thus for the semigroup C*-algebras of Artin–Tits monoids of finite type complement nicely the results in [10, 11, 25] on C*-algebras of right-angled Artin monoids.

The following refers to [40]. Thus $G \equiv G_S$ is the group with finite generating set S, and presentation $\langle S \mid (s,t)_{m_{st}} = (t,s)_{m_{st}}, \text { for } s,t \in S \rangle$, where $(s,t)_m = ststs...$ with m factors, and $M = (m_{st})_{s,t \in S}$ is a symmetric positive integer matrix with $m_{ss} = 1$ and $m_{st} > 1$ if $s \not = t$. Moreover, if W is the group with the same presentation as G but with the additional relations $\{s^2 : s \in S \}$ then W is assumed to be finite. A labeled graph (called the Coxeter diagram of G) is associated to G as follows. S is the set of vertices of the graph, and there is an edge between s and t if $m_{st} > 2$. The edge is labeled $m_{st}$ if $m_{st} > 3$. G is irreducible if this graph is connected. (If $S = S_1 \sqcup S_2$ such that $m_{s_1s_2} = 2$ whenever $s_1 \in S_1$ and $s_2 \in S_2$ then $G_S \cong G_{S_1} \times G_{S_2}$.) We let $P \equiv P_S$ be the monoid given by $\langle S \mid (s,t)_{m_{st}} = (t,s)_{m_{st}} \text { for } s,t \in S \rangle ^+$. As in [40] we let $p : G \rightarrow W$ be the obvious surjection, and $P_{\text {red}} = \{ x \in P : \ell (x) = \ell (p(x)) \}$, where $\ell$ is the length function. For s, $t \in S$ we write $\Delta _{s,t} = (s,t)_{m_{st}}$. We use the following notation from [40]: for g, $h \in P$ we write $g \prec h$ if $h \in gP$ and $h \succ g$ if $h \in Pg$. By Michel [40, Proposition 2.6], for any two elements g, $h \in P$ there is a unique maximal (for $\prec$) element $g \wedge h \in P$ satisfying $g \wedge h \prec g$, h, and as in [40, Section 3] also a unique minimal element $g \vee h \in P$ satisfying g, $h \prec g \vee h$. It follows that P is left reversible and right LCM. For $g \in P$ we let $\mathcal L(g) = \{s \in S : s \prec g \}$ and $\mathcal R(g) = \{s \in S : g \succ s\}$. For $g \in P_{\text {red}}$ and $s \in S$, if $s \not \in \mathcal R(g)$ then $gs \in P_{\text {red}}$, and similarly on the left (see [40, remarks before Proposition 1.5]). A finite sequence $(g_1, \ldots , g_k)$ of elements of $P_{\text {red}} {\setminus } \{1\}$ is called a normal form if $\mathcal R(g_i) \supseteq \mathcal L(g_{i+1})$ for $1 \le i < k$. It is shown in [40, Corollaries 4.2, 4.3] that for each $g \in P$ there is a unique normal form $(g_1, \ldots , g_k)$ such that $g = g_1 \cdots g_k$. We use the notation $\nu (g)$ from [40, p. 367] for the decomposition length of $g \in P$: $\nu (g) = n$ if g has normal form $(g_1, \ldots , g_n)$. If $T \subseteq S$ we write $G_T$, respectively $P_T$, for the Artin–Tits group, respectively monoid, corresponding to $\{ m_{st} : s,t \in T \}$. It is clear that the inclusion $T \subseteq S$ defines a homomorphism $G_T \rightarrow G_S$ carrying $P_T$ to $P_S$.

Lemma 4.10

Let $T \subseteq S$ and let $g \in \langle T \rangle _{G_S}$. Then $\mathcal L(g)$, $\mathcal R(g) \subseteq T$.

Proof

Let $s \in S {\setminus } T$. Suppose that $s \in \mathcal L(g)$. Write $g = t_1 t_2 \cdots t_m$ with $t_i \in T$. Since s, $t_1 \in \mathcal L(g)$ it follows from [40, Proposition 2.5] that $\Delta _{\{s,t_1\}} \prec g$. Then since $t_1 s \prec \Delta _{s,t_1}$ we have $t_1 s \prec g$, and hence $s \prec t_2 \cdots t_m$. Repeating this process we eventually obtain $s \prec t_m$, hence $s = t_m \in T$, contradicting our supposition. The argument for $\mathcal R(g)$ is similar. $\square$

Proposition 4.11

Let $T \subseteq S$. Then the homomorphism $G_T \rightarrow G_S$ is injective.

Proof

Let $g \in P_T$. Then g has a normal form $(g_1, \ldots , g_k)$ in $P_T$. By Lemma 4.10 this is also a normal form in $P_S$. By Michel [40, Corollary 4.3] it follows that $G_T \rightarrow G_S$ is injective on $P_T$. By Michel [40, Corollary 3.2] it follows that $G_T \rightarrow G_S$ is injective. $\square$

As in [40, Section 3], there is a unique element $\Delta \in P_{\text {red}}$ of greatest length (which plays the role of w in Lemma 4.1). We will write $P_0 = P_{\text {red}} {\setminus } \{1,\Delta \}$. Then for $1 \not = g \in P {\setminus } \Delta P$, g has normal form $(g_1, \ldots , g_k)$ with $g_i \in P_0$. We wish to give a normal form for infinite words analogous to the normal form described above for elements of P. For this it is convenient to recall the characterization of $\Omega$ given in [49, Section 7]. For $g \in P$ let $[g] = \{ h \in P : h \prec g \}$. A subset $x \subseteq P$ is hereditary if $[g] \subseteq x$ whenever $g \in x$, and is directed if for all g, $h \in x$, $gP \cap hP \cap x \not = \emptyset$. Then $\Omega$ can be identified with the collection of all nonempty directed hereditary subsets of P. Left reversibility implies that P itself is a directed hereditary subset, corresponding to $\infty \in \Omega$. The finite directed hereditary subsets correspond to the elements of P itself, via $g \in P \leftrightarrow [g] \subseteq P$. Thus $\tilde{\Omega }_{\infty }$ corresponds to $\{x \in \Omega : x \text { is infinite and } x \not = P \}$. For $k \ge 0$ let $X_k = \{ x \in \tilde{\Omega }_{\infty } : \Delta ^k \in x,\ \Delta ^{k+1} \not \in x \}$. Then $\tilde{\Omega }_{\infty } = \sqcup _{k=0}^\infty X_k$.

Lemma 4.12

$X_0$ is in one-to-one correspondence with infinite sequences $(g_1, g_2, \ldots )$ such that $g_i \in P_0$ and $\mathcal R(g_i) \supseteq \mathcal L(g_{i+1})$ for all i. The correspondence pairs such a sequence with the directed hereditary set $\bigcup _{n=1}^\infty [g_1 \cdots g_n]$.

Proof

First we claim that if $x \in X_0$ then x contains a unique element of $P_0$ of maximal length. To see this, first consider two elements g, $h \in P_0 \cap x$. Since x is directed there is $f \in x \cap gP \cap hP$. But then $g \vee h \prec f$. Since x is hereditary, we have $g \vee h \in x$. Since g, $h \in P_0$ we have g, $h \prec \Delta$, and hence $g \vee h \in P_{\text {red}}$. But since $g \vee h \in x$ and $x \in X_0$ we know that $g \vee h \not = \Delta$. Therefore $g \vee h \in P_0 \cap x$. Since $x \cap P_0$ is finite we may apply the preceding argument finitely many times to see that $\bigvee (x \cap P_0)$ is the desired element of maximal length in $x \cap P_0$. Now let $x \in X_0$ and let $g_1$ be the maximal element of $x \cap P_0$. We write $\sigma ^{g_1}(x) = \{ h \in P : g_1 h \in x \} = g_1^{-1} (x \cap g_1 P)$ (with notation borrowed from [49]). It is easy to check that $\sigma ^{g_1}(x) \in X_0$. Let $g_2$ be maximal in $\sigma ^{g_1}(x) \cap P_0$. Then $\mathcal R(g_1) \supseteq \mathcal L(g_2)$: for, if $s \in \mathcal L(g_2) {\setminus } \mathcal R(g_1)$ we may write $g_2 = s g_2'$, and also we have $g_1 s \in P_{\text {red}}$. Since $g_1 s \prec g_1 g_2$ we know that $g_1s \in x$, and hence $g_1 s \in P_0$, contradicting the maximality of $g_1$. Repeating this construction we obtain an infinite sequence $(g_1,g_2, \ldots )$ such that $\mathcal R(g_i) \supseteq \mathcal L(g_{i+1})$ for all i. Since $g_1 \cdots g_n \in x$ for all n, we have $\bigcup _{n=1}^\infty [g_1 \cdots g_n] \subseteq x$. Conversely, let $h \in x$. We want to show that $h \in \bigcup _{n=1}^\infty [g_1 \cdots g_n]$. First note that if $p \in P_\mathrm{red}$ and $q \in P$ has normal form of length n, i.e., of the form $(q_1, \dotsc , q_n)$, then $p \vee q$ must have normal form of length n. This is straightforward using induction on n: For $n=1$, this follows from the fact that $P_\mathrm{red}$ is closed under $\vee$. Assume that our claim is true for $n-1$. Write $p \vee q_1 = q_1 r$ with $r \in P_\mathrm{red}$. Then $p \vee q = (p \vee q_1) \vee q = (q_1 r) \vee q = q_1 (r \vee (q_2 \cdots q_n))$. By the induction hypothesis, $r \vee (q_2 \cdots q_n)$ has normal form of length $n-1$, so that $q_1 (r \vee (q_2 \cdots q_n))$ has normal form of length at most n. The length must then be equal to n since q has normal form of length n. Now consider $h \in x$ as above. Suppose that h has normal form of length n. Then we claim that there exists $r \in P$ such that $hr = g_1 \cdots g_n$. We prove this inductively on n. Our claim is obviously true for $n=1$. Assume it is true for $n-1$. By the above observation, we know that $g_1 \vee h$ has normal form of length n, say $(g_1, q_2, \dotsc , q_n)$. Hence there exists $s \in P$ such that $hs = g_1 q_2 \cdots q_n$. Applying the induction hypothesis to $q_2 \cdots q_n$ and $\sigma ^{g_1}(x)$, we obtain $t \in P$ such that $q_2 \cdots q_n t = g_2 \cdots g_n$. Then set $r = st$. We have $hr = hst = g_1 q_2 \cdots q_n t = g_1 g_2 \cdots g_n$. This proves our claim. Hence it follows that $h \prec g_1 \cdots g_n$, and thus $h \in \bigcup _{n=1}^\infty [g_1 \cdots g_n]$. Therefore, $x \subseteq \bigcup _{n=1}^\infty [g_1 \cdots g_n]$, as desired.    We have shown that for each $x \in X_0$ there is an infinite sequence $(g_1, g_2, \ldots )$ such that $\mathcal R(g_i) \supseteq \mathcal L(g_{i+1})$ for all i and $x = \bigcup _{n=1}^\infty [g_1 \cdots g_n]$. If $(h_1, h_2, \ldots )$ is a sequence in $P_0$ such that $\mathcal R(h_i) \supseteq \mathcal L(h_{i+1})$ for all i and $x = \bigcup _{n=1}^\infty [h_1 \cdots h_n]$, then $g_1 = h_1$ since both equal the unique maximal element of $x \cap P_0$. Inductively, $g_{i+1} = h_{i+1}$ is the unique maximal element of $\sigma ^{g_1 \cdots g_i}(x) \cap P_0$. Therefore the sequence $(g_1,g_2, \ldots )$ associated to x is unique. Finally, if $(g_1, g_2, \ldots )$ is any sequence in $P_0$ such that $\mathcal R(g_i) \supseteq \mathcal L(g_{i+1})$ for all i then $\Delta \not \in \bigcup _{n=1}^\infty [g_1 \cdots g_n]$. Therefore $(g_1, g_2, \ldots )$ is associated to the directed hereditary set $\bigcup _{n=1}^\infty [g_1 \cdots g_n]$ in $X_0$. $\square$

Remark 4.13

The above proof actually shows that $\tilde{\Omega }_{\infty }$ is in one-to-one correspondence with sequences $(g_1,g_2,\ldots )$ such that $g_i \in P_{\text {red}} {\setminus }\{1\}$, $\mathcal R(g_i) \supseteq \mathcal L(g_{i+1})$ for all i, and $g_i \in P_0$ eventually.

Remark 4.14

We extend $\mathcal L$ to $\tilde{\Omega }_{\infty }$ by setting $\mathcal L(x) := \mathcal L(g_1)$ if x has infinite normal form $(g_1, g_2, \ldots )$. For $g \in P$ we let $Z(g) = \{x \in \tilde{\Omega }_{\infty } : g \in x \}$. We note that for g, $h \in P$ we have $Z(g) \cap Z(h) \cap X_0 \not = \emptyset$ if and only if $g \vee h \not \in \Delta P$. In this case $Z(g) \cap Z(h) \cap X_0 = Z(g \vee h) \cap X_0$. The collection $\{Z(g) {\setminus } \bigcup _{j=1}^n Z(h_j) : h_j \in gP \text { for } 1 \le j \le n \}$ is a base of compact-open sets for the topology of $\tilde{\Omega }_{\infty }$. Then $X_0$ is a compact-open subset of $\tilde{\Omega }_{\infty }$ that is transversal, and hence the groupoid $G \ltimes \tilde{\Omega }_{\infty }$ is equivalent to its restriction to $X_0$.

Proposition 4.15

Let G be an Artin–Tits group of finite type with at least three generators, and suppose that there exist s, t in S with $2< m_{st} < \infty$. Then $G \ltimes \tilde{\Omega }_{\infty }$ is not amenable.

Proof

Let $G_{st}$ be the subgroup generated by s and t. Then $G_{st}$ is nonamenable. Let $x = (\Delta _{s,t}, \Delta _{s,t}, \ldots ) \in X_0$. Then the isotropy of $G \ltimes \tilde{\Omega }_{\infty }$ at x contains $G_{st}$, and hence is nonamenable. Since there is a point with nonamenable isotropy, the groupoid is nonamenable. $\square$

We now give several lemmas on the existence of certain normal forms. Note that since $\Delta$ is the unique element of $P_{\text {red}}$ of maximal length, $s \in \mathcal L(\Delta )$ and $s \in \mathcal R(\Delta )$ for all $s \in S$. Moreover $\Delta$ is the only element of $P_{\text {red}}$ with this property: if $g \in P_{\text {red}}$ is such that $s \prec g$ for all $s \in S$, then $\ell (p(s)p(g)) = \ell (p(g)) - 1 < \ell (p(g))$ for all $s \in S$, so p(g) must have maximal length in W. Therefore $g = \Delta$ (and a similar argument works on the right). If $T \subseteq S$ we write $\Delta _T$ for the unique element in $P_{T,\text {red}}$ of maximal length ($P_{T,\text {red}} \subseteq P_{\text {red}}$ by Proposition 4.11).

Lemma 4.16

Let $T \subseteq S$ and $g \in P_{\text {red}}$ be such that $T \subseteq \mathcal L(g)$. Then $\Delta _T \prec g$.

Proof

Let $M = \{h \in P : h \in \langle T \rangle ^+ \text { and } h \prec g \}$. Since $g \in P_{\text {red}}$ we know that $M \subseteq P_{\text {red}}$, and so M is finite. We verify the two conditions in [40, Lemma 1.4]. For the first condition, let $h \in M$ and $k \in P$ be such that $k \prec h$. By Proposition 4.11 we have $k \in P_T$, and hence $k \in M$. For the second condition, let $k \in P$ and s, $t \in S$ such that ks, $kt \in M$. By the first condition $k \in M \subseteq P_T$. Therefore s, $t \in G_T$, hence s, $t \in T$. Since $k \in M$ we have $k \prec g$, so $g = k g'$. Then ks, $kt \prec kg'$, so s, $t \prec g'$. By Michel [40, Proposition 1.5] we have $\Delta _{s,t} \prec g'$, and hence $k \Delta _{s,t} \prec g$. Therefore $k \Delta _{s,t} \in M$. Now by Michel [40, Lemma 1.4] there is $f \in M$ such that $M = \{ h \in P : h \prec f \}$. Then $f \in P_{T,\text {red}}$. Since $T \subseteq M$ we have $t \prec f$ for all $t \in T$. Then $tf \not \in P_{T,\text {red}}$ for all $t \in T$, so f is maximal in $P_{T,\text {red}}$. Therefore $f = \Delta _T$. $\square$

Note that we obtain an alternative proof for Lemma 4.16 by observing that the canonical map $P_T \rightarrow P_S$ is $\vee$-preserving. This, in turn, follows for instance from the recipe for computing $\vee$ using right reversing (see [21, Proposition 6.10]), which does not depend on the ambient monoid.

Lemma 4.17

Let T, $U \subseteq S$ be such that $m_{tu} = 2$ for all $t \in T$ and $u \in U$ (in particular, T and U are disjoint). Let $g \in P_T$. Then $\mathcal L(g \Delta _U) = \mathcal L(g) \cup U$ and $\mathcal R(g \Delta _U) = \mathcal R(g) \cup U$.

Proof

The containments $\supseteq$ are clear. Let $s \in \mathcal L(g \Delta _U)$ and suppose that $s \not \in U$. Then since $U \subseteq \mathcal L(g \Delta _U)$, Lemma 4.16 implies that $\Delta _{U \cup \{s\}} \prec g \Delta _U = \Delta _U g$, hence $\Delta _U s \prec \Delta _U g$, and hence $s \prec g$. Therefore $s \in \mathcal L(g)$ as required. A similar argument on the right gives the second statement of the lemma. $\square$

For the next several lemmas we will consider a portion of the Coxeter diagram that is linear, i.e. a tree with no vertex having valence greater than two.

Lemma 4.18

Let $a_1, \ldots , a_p \in S$ be such that $m_{a_\mu ,a_{\mu +1}} > 2$ for $1 \le \mu < p$ and $m_{a_\mu ,a_\nu } = 2$ if $|\mu - \nu | \ge 2$ (so that $a_1, \ldots , a_p$ forms a linear subgraph of the Coxeter diagram of G). For $1 \le \mu \le p$ we have

Moreover the above elements are in $P_0$.

Proof

It is clear that the elements are in $P_0$, and that in all parts the containments $\supseteq$ hold. We prove the reverse containments.

For (i) suppose that $a_{\mu -1} \in \mathcal L(a_\mu \Delta _{\{a_1,\ldots ,a_{\mu -1}\}} \Delta _{\{a_{\mu +1},\ldots , a_p\}})$. Then $a_\mu$, $a_{\mu -1} \in \mathcal L(a_\mu \Delta _{\{a_1,\ldots ,a_{\mu -1}\}} \Delta _{\{a_{\mu +1},\ldots , a_p\}})$, so by Michel [40, Proposition 2.5] we have $a_\mu a_{\mu -1} a_\mu \prec \Delta _{\{a_{\mu -1},a_\mu \}} \prec a_\mu \Delta _{\{a_1,\ldots ,a_{\mu -1}\}} \Delta _{\{a_{\mu +1},\ldots , a_p\}}$. But then $a_{\mu -1} a_\mu \prec \Delta _{\{a_1,\ldots ,a_{\mu -1}\}} \Delta _{\{a_{\mu +1},\ldots , a_p\}}$, which contradicts the fact that $a_\mu \not \in G_{\{a_\nu : \nu \not = \mu \}}$ (by Proposition 4.11). A similar argument shows that $a_{\mu +1} \not \in \mathcal L(a_\mu \Delta _{\{a_1,\ldots ,a_{\mu -1}\}} \Delta _{\{a_{\mu +1},\ldots , a_p\}})$. For (ii) suppose $a_\mu \in \mathcal R(a_\mu \Delta _{\{a_1,\ldots ,a_{\mu -1}\}} \Delta _{\{a_{\mu +1},\ldots , a_p\}})$. Then $a_1$, $\ldots$, $a_p \in \mathcal R(a_\mu \Delta _{\{a_1,\ldots ,a_{\mu -1}\}} \Delta _{\{a_{\mu +1},\ldots , a_p\}})$. By Lemma 4.16 we have $a_\mu \Delta _{\{a_1,\ldots ,a_{\mu -1}\}} \Delta _{\{a_{\mu +1},\ldots , a_p\}} \succ \Delta _{\{a_1,\ldots ,a_p\}}$. By Michel [40, Proposition 3.1], applied to $G_{\{a_1,\ldots ,a_p\}}$, we have $\Delta _{\{a_1,\ldots ,a_p\}} \prec a_\mu \Delta _{\{a_1,\ldots ,a_{\mu -1}\}} \Delta _{\{a_{\mu +1},\ldots , a_p\}}$, contradicting (i). The proof of (iii) is similar to that of (ii), and the proof of (iv) is similar to that of (i). For (v), first suppose that $a_{\mu -2} \in \mathcal L(a_{\mu -1} \Delta _{\{a_1,\ldots ,a_{\mu -2}\}} \Delta _{\{a_{\mu +1},\ldots ,a_p\}} a_\mu )$. Then $a_{\mu -2}$, $a_{\mu -1} \in \mathcal L(a_{\mu -1} \Delta _{\{a_1,\ldots ,a_{\mu -2}\}} \Delta _{\{a_{\mu +1},\ldots ,a_p\}} a_\mu )$. Now the proof is similar to that of (i). Next suppose that $a_\mu \in \mathcal L(a_{\mu -1} \Delta _{\{a_1,\ldots ,a_{\mu -2}\}} \Delta _{\{a_{\mu +1},\ldots ,a_p\}} a_\mu )$. As before we then have $a_{\mu -1} a_\mu a_{\mu -1} \prec a_{\mu -1} \Delta _{\{a_1,\ldots ,a_{\mu -2}\}} \Delta _{\{a_{\mu +1},\ldots ,a_p\}} a_\mu$, and hence $a_\mu a_{\mu -1} \prec \Delta _{\{a_1,\ldots ,a_{\mu -2}\}} \Delta _{\{a_{\mu +1},\ldots ,a_p\}} a_\mu$, and hence $a_\mu \in \mathcal L(\Delta _{\{a_1,\ldots ,a_{\mu -2}\}} \Delta _{\{a_{\mu +1},\ldots ,a_p\}} a_\mu )$. By Lemma 4.17 we then have $a_\mu \in \mathcal L(\Delta _{\{a_{\mu +1},\ldots ,a_p\}} a_\mu )$, contradicting (iii) (applied to the subset $\{a_\mu ,\ldots ,a_p\}$). The proof of (vi) is similar to that of (v). Finally, it is clear that, for example, $\Delta _{\{a_1,\ldots ,a_{\mu -1}\}}\Delta _{\{a_{\mu +1},\ldots , a_p\}} \in P_{\{a_1,\ldots ,a_p\} {\setminus } \{a_\mu \},\text {red}}$ and that $a_\mu \not \in \mathcal L(\Delta _{\{a_1,\ldots ,a_{\mu -1}\}} \Delta _{\{a_{\mu +1},\ldots , a_p\}})$. Therefore $a_\mu \Delta _{\{a_1,\ldots ,a_{\mu -1}\}} \Delta _{\{a_{\mu +1},\ldots , a_p\}} \in P_{\{a_1,\ldots ,a_p\},\text {red}}$. Similar arguments demonstrate the same for parts (ii)–(vi). $\square$

Definition 4.19

Let $T \subseteq S$. For $\emptyset \not = T_1$, $T_2 \subsetneq T$ we write $T_1 \sim _T T_2$ if there is $g \in P_T$ with normal form $(g_1, \ldots , g_j)$ such that $\mathcal L(g_1) = T_1$ and $\mathcal R(g_j) = T_2$.

Remark 4.20

It is clear that $\sim _T$ is an equivalence relation on the proper nonempty subsets of T. (T cannot be $\sim _T$-equivalent to a proper subset by Michel [40, Proposition 3.1].)

Lemma 4.21

If T determines a connected portion of the Coxeter diagram of G then $\{s\} \sim _T \{t\}$ for all s, $t \in T$.

Proof

Let s, $t \in T$. Since T is connected there are $s_1$, $\ldots$, $s_k \in T$ defining a linear subgraph of the Coxeter diagram (as in Lemma 4.18) such that $s = s_1$ and $t = s_k$. Let $g = s_1 s_2^2 s_3^2 \cdots s_{k-1}^2 s_k$. Then g has normal form $(s_1 s_2, s_2 s_3, \ldots , s_{k-1} s_k)$, and hence $\mathcal L(g_1) = \mathcal L(s_1 s_2) = s_1$ and $\mathcal R(g_{k-1}) = \mathcal R(s_{k-1}s_k) = s_k$. $\square$

Lemma 4.22

Let $T \subseteq U \subseteq S$. Then $\sim _T \subseteq \sim _U$.

Proof

This follows from Lemma 4.10. $\square$

Remark 4.23

By Lemma 4.22, equivalences relative to a subgraph still hold relative to a larger subgraph. We will use this as needed without further mention.

In the following Lemmas 4.24–4.29, let $T = \{s_1,\ldots ,s_k\} \subseteq S$ be a linear subgraph of the Coxeter diagram of G (as in Lemma 4.18).

Lemma 4.24

Let $1< i < k$. Then $\{s_1,\ldots ,s_{i-1}\} \sim _{\{s_1,\ldots ,s_{i+1}\}} \{s_1,\ldots ,s_i\}$.

Proof

Letting $\{s_1,\ldots ,s_i\}$ play the role of $\{a_1,\ldots ,a_p\}$ in Lemma 4.18 (iii) and (iv), and letting $\mu = i$, we have $\mathcal L(\Delta _{\{s_1,\ldots ,s_{i-1}\}}s_i) = \{s_1,\ldots ,s_{i-1}\}$ and $\mathcal R(\Delta _{\{s_1,\ldots ,s_{i-1}\}}s_i) = \{s_1,\ldots ,s_{i-2},s_i\}$ (or $\{s_i\}$ if $i=2$). Next, letting $\{s_1,\ldots ,s_{i+1}\}$ play the role of $\{a_1,\ldots ,a_p\}$ in Lemma 4.18 (v) and (vi), and letting $\mu = i+1$, we have $\mathcal L(s_i \Delta _{\{s_1,\ldots ,s_{i-1}\}}s_{i+1}) = \{s_1,\ldots ,s_{i-2},s_i\}$ (or $\{s_i\}$ if $i=2$) and $\mathcal R(s_i \Delta _{\{s_1,\ldots ,s_{i-1}\}}s_{i+1}) = \{s_1,\ldots ,s_{i-1},s_{i+1}\}$. Finally, letting $\{s_1,\ldots ,s_{i+1}\}$ play the role of $\{a_1,\ldots ,a_p\}$ in Lemma 4.18 (i) and (ii), and letting $\mu = i+1$, we have $\mathcal L(s_{i+1} \Delta _{\{s_1,\ldots ,s_i\}}) = \{s_1,\ldots ,s_{i-1},s_{i+1}\}$ and $\mathcal R(s_{i+1} \Delta _{\{s_1,\ldots ,s_i\}}) = \{s_1,\ldots ,s_i\}$. Let $g_1 = \Delta _{\{s_1,\ldots ,s_{i-1}\}}s_i$, $g_2 = s_i \Delta _{\{s_1,\ldots ,s_{i-1}\}}s_{i+1}$, and $g_3 = s_{i+1} \Delta _{\{s_1,\ldots ,s_i\}}$. Then $g = g_1g_2g_3$ has normal form $(g_1,g_2,g_3)$, and $\mathcal L(g_1) = \{s_1,\ldots ,s_{i-1}\}$, $\mathcal R(g_3) = \{s_1,\ldots ,s_i\}$. $\square$

Lemma 4.25

Let $1< i < k$. Then $\{s_i,\ldots ,s_k\} \sim _{\{s_{i-1},\ldots ,s_k\}} \{s_{i+1},\ldots ,s_k\}$.

Proof

The proof is analogous to the proof of Lemma 4.24. $\square$

Lemma 4.26

Let $1< i < k$. Then $\{s_1,s_{i+1},\ldots ,s_k\} \sim _{\{s_1,s_i,\ldots ,s_k\}} \{s_1,s_k\}$.

Proof

Applying Lemma 4.25 repeatedly we find that $\{s_{i+1},\ldots ,s_k\} \sim _{\{s_i,\ldots ,s_k\}} \{s_k\}$. Suppose that $i > 2$. Then in fact we have $\{s_{i+1},\ldots ,s_k\} \sim _{\{s_3,\ldots ,s_k\}} \{s_k\}$. By Lemma 4.17 we then have $\{s_1,s_{i+1},\ldots ,s_k\} \sim _{\{s_1,s_i,\ldots ,s_k\}} \{s_1,s_k\}$. For the case where $i = 2$, we consider the proof of Lemma 4.25 for this case. For that we start with the normal form $(\Delta _{\{s_3,\ldots ,s_k\}}s_2, s_2 \Delta _{\{s_4,\ldots ,s_k\}}s_3, s_3 \Delta _{\{s_4,\ldots ,s_k\}})$, for which $\mathcal L(\cdot _1) = \{s_3,\ldots ,s_k\}$ and $\mathcal R(\cdot _3) = \{s_4,\ldots ,s_k\}$. We modify this to include $s_1$ as follows. It is only in the first two terms of the normal form that $s_2$ occurs, obstructing the use of Lemma 4.17. By Lemma 4.18 (v) and (vi), with $\{s_1,\ldots ,s_k\}$ playing the role of $\{a_1,\ldots ,a_p\}$ and $\mu = 2$, we have $\mathcal L(s_1 \Delta _{\{s_3,\ldots ,s_k\}} s_2) = \{s_1,s_3,\ldots ,s_k\}$ and $\mathcal R(s_1 \Delta _{\{s_3,\ldots ,s_k\}} s_2) = \{s_2,s_4,\ldots ,s_k\}$. By Lemma 4.18 (v) and (vi), with $\{s_1,\ldots ,s_k\}$ playing the role of $\{a_1,\ldots ,a_p\}$ and $\mu = 3$, we have $\mathcal L(s_2 s_1 \Delta _{\{s_4,\ldots ,s_k\}} s_3) = \{s_2,s_4,\ldots ,s_k\}$ and $\mathcal R(s_2 s_1 \Delta _{\{s_4,\ldots ,s_k\}} s_3) = \{s_1,s_3,s_5,\ldots ,s_k\}$. By Lemmas 4.17, and 4.18 (i) and (ii) with $\{s_3,\ldots ,s_k\}$ playing the role of $\{a_1,\ldots ,a_p\}$ and $\mu = 1$, we have $\mathcal L(s_1 s_3 \Delta _{\{s_4,\ldots ,s_k\}}) = \{s_1\} \cup \mathcal L(s_3 \Delta _{\{s_4,\ldots ,s_k\}}) = \{s_1,s_3,s_5,\ldots ,s_k\}$ and $\mathcal R(s_1 s_3 \Delta _{\{s_4,\ldots ,s_k\}}) = \{s_1\} \cup \mathcal R(s_3 \Delta _{\{s_4,\ldots ,s_k\}}) = \{s_1,s_4,\ldots ,s_k\}$. So $(s_1 \Delta _{\{s_3,\ldots ,s_k\}} s_2, s_2 s_1 \Delta _{\{s_4,\ldots ,s_k\}} s_3, s_1 s_3 \Delta _{\{s_4,\ldots ,s_k\}})$ is a normal form with $\mathcal L(\cdot _1) = \{s_1,s_3,\ldots ,s_k\}$ and $\mathcal R(\cdot _3) = \{s_1,s_4,\ldots ,s_k\}$. If we combine this with the equivalence at the beginning of the proof we are finished. $\square$

Lemma 4.27

Let $1< i < k$. Then $\{s_1,s_{i+1},\ldots ,s_k\} \sim _T T {\setminus } \{s_i\}$.

Proof

By repeated application of Lemma 4.24 we have that $\{s_1\} \sim _{\{s_1,\ldots ,s_{i-1}\}} \{s_1,\ldots ,s_{i-2}\}$. Then by Lemma 4.17, multiplication by $\Delta _{\{s_{i+1},\ldots ,s_k\}}$ gives

$$\begin{aligned} \{s_1,s_{i+1},\ldots ,s_k\} \sim _{T {\setminus } \{s_i\}} \{s_1,\ldots ,s_{i-2},s_{i+1},\ldots ,s_k\}. \end{aligned}$$

(4)

From Lemma 4.18 (iii) and (iv), with $\{s_1,\ldots ,s_{i-1}\}$ playing the role of $\{a_1,\ldots ,a_p\}$ and $\mu = i-1$, we find that $\{s_1,\ldots ,s_{i-2}\} \sim _{\{s_1,\ldots ,s_{i-1}\}} \{s_1,\ldots ,s_{i-3},s_{i-1}\}$. Then by Lemma 4.17, multiplication by $\Delta _{\{s_{i+1},\ldots ,s_k\}}$ gives

$$\begin{aligned} \{s_1,\ldots ,s_{i-2},s_{i+1},\ldots ,s_k\} \sim _{\{s_1,\ldots ,s_{i-1},s_{i+1},\ldots ,s_k\}} \{s_1,\ldots ,s_{i-3},s_{i-1},s_{i+1},\ldots ,s_i\}. \end{aligned}$$

(5)

By Lemma 4.18 (v) and (vi), with $\{s_1,\ldots ,s_k\}$ playing the role of $\{a_1,\ldots ,a_p\}$ and $\mu = i$, we have

$$\begin{aligned} \{s_1,\ldots ,s_{i-3},s_{i-1},s_{i+1},\ldots ,s_k\} \sim _T \{s_1,\ldots ,s_{i-2},s_i,s_{i+2},\ldots ,s_k\}. \end{aligned}$$

(6)

By Lemma 4.18 (i) and (ii), with $\{s_1,\ldots ,s_k\}$ playing the role of $\{a_1,\ldots ,a_p\}$ and $\mu = i$, we have

$$\begin{aligned} \{s_1,\ldots ,s_{i-2},s_i,s_{i+2},\ldots ,s_k\} \sim _T \{s_1,\ldots ,s_{i-1},s_{i+1},\ldots ,s_k\}. \end{aligned}$$

(7)

The combination of equivalences (4)–(7) proves the lemma. $\square$

Lemma 4.28

We have $\{s_1\} \sim _T \{s_1,s_k\}$.

Proof

We know that $\mathcal L(s_1s_2) = \{s_1\}$ and $\mathcal R(s_1s_2) = \{s_2\}$. By Lemma 4.18 (i) and (ii), with $\{s_1,s_2,s_3\}$ playing the role of $\{a_1,\ldots ,a_p\}$ and $\mu = 2$, we have $\mathcal L(s_2s_1s_3) = \{s_2\}$ and $\mathcal R(s_2s_1s_3) = \{s_1,s_3\}$. For $i > 2$ we know that $\mathcal L(s_is_{i+1}) = \{s_i\}$ and $\mathcal R(s_is_{i+1}) = \{s_{i+1}\}$. Then by Lemma 4.17 we have $\mathcal L(s_1s_is_{i+1}) = \{s_1,s_i\}$ and $\mathcal R(s_1s_is_{i+1}) = \{s_1,s_{i+1}\}$. Combining these gives the lemma. $\square$

Lemma 4.29

Let $1 \le i, j \le k$. Then $\{s_j\} \sim _T T {\setminus } \{s_i\}$.

Proof

First suppose $1 < i$. By Lemma 4.21 we have $\{s_j\} \sim _T \{s_1\}$. By Lemma 4.28 we have $\{s_1\} \sim _T \{s_1,s_k\}$. By Lemma 4.26 we have $\{s_1,s_k\} \sim _T \{s_1,s_{i+1},\ldots ,s_k\}$. By Lemma 4.27 we have $\{s_1,s_{i+1},\ldots ,s_k\} \sim _T T {\setminus } \{s_i\}$. Combining these equivalence gives the desired result. For the case $i = 1$, note that repeated use of Lemma 4.25 gives $\{s_2,\ldots ,s_k\} \sim _T \{s_k\}$. Combining this with Lemma 4.21 gives $\{s_j\} \sim _T \{s_2,\ldots ,s_k\}$. $\square$

If G is irreducible and of finite type then the Coxeter diagram must be a tree with at most one vertex having valence greater than two; if there is such a vertex it has valence three. Moreover, in the case that there is a vertex of valence three, let $v \in S$ be the vertex of valence three, and let the three linear pieces of $S {\setminus } \{v\}$ have $n_1$, $n_2$, $n_3$ vertices, respectively. Then we may assume that $n_3 = 1$ and that $n_2 = 1$ or 2 ([5, Chapter VI, Section 4.1, Théorème 1]).

Remark 4.30

We next will generalize the previous lemmas to the case where the Coxeter diagram is a tree of this type. Thus in the following Lemmas 4.31–4.37, we will assume that $S = \{s_1,\ldots ,s_k,u\}$ with $k \ge 3$, that $m_{s_i,s_{i+1}} > 2$ for $1 \le i < k$, that $m_{s_\ell ,u} > 2$ for $\ell = 2$ or 3, that $\ell < k$, and that $m_{t,t'} = 2$ for all other pairs of elements of S.

Lemma 4.31

In Lemma 4.18 let $\{s_1,\ldots ,s_k\}$ play the role of $\{a_1,\ldots ,a_p\}$. The elements of $P_{\text {red}}$ in Lemma 4.18 may be modified so that u is included in all of their $\mathcal L$ and $\mathcal R$ sets.

Proof

We indicate the procedure for the first element; the other two are managed similarly. Let $1 \le i \le k$ and consider the element $s_i \Delta _{\{s_1,\ldots ,s_{i-1}\}} \Delta _{\{s_{i+1},\ldots ,s_k\}}$. There are three cases depending on the relation between i and $\ell$. Recall that $m_{s_j,u} > 2$ if and only if $j = \ell$.

Case 1: $\ell < i$. We consider the element $g = s_i \Delta _{\{u,s_1,\ldots ,s_{i-1}\}} \Delta _{\{s_{i+1},\ldots ,s_k\}}$.

Case 2: $\ell > i$. We consider the element $g = s_i \Delta _{\{s_1,\ldots ,s_{i-1}\}} \Delta _{\{u,s_{i+1},\ldots ,s_k\}}$.

Case 3: $\ell = i$. We consider the element $g = \Delta _{\{u,s_i\}} \Delta _{\{s_1,\ldots ,s_{i-1}\}} \Delta _{\{s_{i+1},\ldots ,s_k\}}$.

In all three cases we claim that $\mathcal L(g) = \{u\} \cup \{s_j : j \not = i-1,\ i+1 \}$ and $\mathcal R(g) = \{u\} \cup \{s_j : j \not = i \}$. The containments $\supseteq$ are clear. The proofs of the containments $\subseteq$ are identical to those in the proof of Lemma 4.18 (i) and (ii). An analogous argument works for parts (iii)–(vi) of Lemma 4.18. $\square$

Lemma 4.32

For $1 \le i < j \le k$, $\{u,s_i\} \sim _S \{u,s_j\}$.

Proof

If $\ell < i$ or $\ell > j$ we have the normal form $(us_is_{i+1}, us_{i+1}s_{i+2}, \ldots , us_{j-1}s_j)$; the $\mathcal L$ and $\mathcal R$ sets of the terms follow from Lemma 4.17. If $i< \ell < j$ we have the normal form $(us_is_{i+1}, \ldots , u s_{\ell -2}s_{\ell -1}, us_{\ell -1}s_\ell , s_\ell s_{\ell +1} u, \ldots , u s_{j-1}s_j)$; the $\mathcal L$ and $\mathcal R$ sets of the terms follow from Lemma 4.17 except for the two terms involving $s_\ell$. For $us_{\ell - 1}s_\ell$ we may use Lemma 4.18 (iii) and (iv) (with $a_1 = u$, $a_2 = s_\ell$, $a_3 = s_{\ell - 1}$, and $\mu = 2$), while for $s_\ell s_{\ell + 1} u$ we may use Lemma 4.18 (iii) and (iv) (with $a_1 = u$, $a_2 = s_\ell$, $a_3 = s_{\ell + 1}$, and $\mu = 1$). If $\ell = i$ we use the normal form $(\Delta _{ \{ u,s_\ell \} } s_{\ell + 1}, u s_{\ell + 1} s_{\ell + 2}, \cdots , u s_{j-1} s_j)$, where for the first term we use Lemma 4.18 (iii) and (iv) (with $a_1 = u$, $a_2 = s_\ell$, $a_3 = s_{\ell + 1}$, and $\mu = 3$). If $\ell = j$ we use the normal form $(u s_i s_{i+1}, \ldots , u s_{\ell -2} s_{\ell -1}, s_{\ell - 1} \Delta _{ \{u, s_\ell \} })$, where for the last term we use Lemma 4.18 (i) and (ii) (with $a_1 = s_{\ell - 1}$, $a_2 = s_\ell$, $a_3 = u$, and $\mu = 1$). $\square$

Lemma 4.33

We have $\{u,s_1\} \sim _S \{u,s_1,s_k\}$.

Proof

If $\ell = 2$ we have the normal form $(us_1s_2, s_2s_1s_3u, us_1s_3s_4, \ldots us_1s_{k-1}s_k)$: the $\mathcal L$ and $\mathcal R$ sets of the first term are calculated using Lemma 4.18, for the second term the calculation is similar to the arguments in the proof of that lemma, and for the remaining terms Lemma 4.17 applies. If $\ell = 3$ we have the normal form $(us_1s_2, us_2s_1s_3, s_1s_3s_4u, us_1s_4s_5, \ldots , us_1s_{k-1}s_k)$. The arguments here are analogous. The cases $\ell = 2$, $k = 3$ and $\ell = 3$, $k = 4$ are slightly different. When $\ell = 2$ and $k = 3$ we use the normal form $(u s_1 s_2, s_2 s_1 s_3 u)$. The calculation of $\mathcal L(s_2 s_1 s_3 u)$ and $\mathcal R(s_2 s_1 s_3 u)$ are as in the proof of Lemma 4.18. When $\ell = 3$ and $k = 4$ we use the normal form $(u s_1 s_2, u s_2 s_1 s_3, s_1 s_3 s_4 u)$; again, calculation of the relevant $\mathcal L$ and $\mathcal R$ sets is as in the proof of Lemma 4.18. $\square$

Lemma 4.34

Let $1< i < k$. Then $\{u,s_1,\ldots ,s_{i-1}\} \sim _{\{u,s_1,\ldots ,s_{i+1}\}} \{u,s_1,\ldots ,s_i\}$.

Proof

This is the same as Lemma 4.24 with u included in all $\mathcal L$ and $\mathcal R$ sets. Since the proof of Lemma 4.24 used only Lemma 4.18, an analogous proof can be given here using Lemma 4.31. $\square$

Lemma 4.35

Let $1< i < k$. Then $\{u,s_i,\ldots ,s_k\} \sim _{\{u,s_{i-1},\ldots ,s_k\}} \{u,s_{i+1},\ldots ,s_k\}$.

Proof

This is the same as Lemma 4.25 with u included in all $\mathcal L$ and $\mathcal R$ sets. The proof is analogous to that of Lemma 4.34. $\square$

Lemma 4.36

Let $1< i < k$. Then $\{u,s_1,s_{i+1},\ldots ,s_k\} \sim _S \{u,s_1,s_k\}$.

Proof

Applying Lemma 4.35 repeatedly we find that $\{u,s_{i+1},\ldots ,s_k\} \sim _{\{u,s_i,\ldots ,s_k\}} \{u,s_k\}$. As in the proof of Lemma 4.26, if $i > 2$ then Lemma 4.17 finishes the proof. Suppose $i = 2$. First consider the case $\ell = 2$. We start with the normal form from Lemma 4.26: $(\Delta _{\{s_3,\ldots ,s_k\}}s_2, s_2 \Delta _{\{s_4,\ldots ,s_k\}}s_3, s_3 \Delta _{\{s_4,\ldots ,s_k\}})$. As in Lemma 4.31 we have the modified version $(\Delta _{\{s_3,\ldots ,s_k\}}\Delta _{\{u,s_2\}}, \Delta _{\{u,s_2\}} \Delta _{\{s_4,\ldots ,s_k\}}s_3, us_3 \Delta _{\{s_4,\ldots ,s_k\}})$. Now, imitating the proof of Lemma 4.26 we claim that there is a normal form $(s_1\Delta _{\{s_3,\ldots ,s_k\}}\Delta _{\{u,s_2\}}, \Delta _{\{u,s_2\}} s_1 \Delta _{\{s_4,\ldots ,s_k\}}s_3, us_1s_3 \Delta _{\{s_4,\ldots ,s_k\}})$. We have

$$\begin{aligned} \mathcal L(s_1\Delta _{\{s_3,\ldots ,s_k\}}\Delta _{\{u,s_2\}})&= \{u,s_1,s_3,\ldots ,s_k\}, \text { by Lemma 4.31} \\ \mathcal R(s_1\Delta _{\{s_3,\ldots ,s_k\}}\Delta _{\{u,s_2\}})&= \{u,s_2,s_4,\ldots ,s_k\}, \text { by Lemma 4.31} \\ \mathcal L(\Delta _{\{u,s_2\}} s_1 \Delta _{\{s_4,\ldots ,s_k\}}s_3)&= \{u,s_2,s_4,\ldots ,s_k\}, \text { by Lemma 4.31} \\ \mathcal R(\Delta _{\{u,s_2\}} s_1 \Delta _{\{s_4,\ldots ,s_k\}}s_3)&= \{u,s_1,s_3,s_5,\ldots ,s_k\}, \text { by Lemma 4.31} \\ \mathcal L(us_1s_3 \Delta _{\{s_4,\ldots ,s_k\}})&= \{u,s_1,s_3,s_5,\ldots ,s_k\} \\ \mathcal R(us_1s_3 \Delta _{\{s_4,\ldots ,s_k\}})&= \{u,s_1,s_4,\ldots ,s_k\},\end{aligned}$$

where to justify the last two equalities we apply Lemma 4.31 to $us_3 \Delta _{\{s_4,\ldots ,s_k\}}$, and then Lemma 4.17 to incude $s_1$. In the case $\ell = 3$ we use the normal form $(s_1 \Delta _{\{u, s_3, \ldots , s_k\}} s_2, s_2 s_1 \Delta _{\{ s_4, \ldots , s_k \}} \Delta _{\{s_3,u\}}, s_1 \Delta _{\{u,s_3\}} \Delta _{\{s_4, \ldots , s_k\}})$. The $\mathcal L$ and $\mathcal R$ sets of the first term are verified using Lemma 4.31, of the third term using Lemmas 4.31 and 4.17, and of the second term by arguments analogous to those in the proof of Lemma 4.18. $\square$

Lemma 4.37

Let $1< i < k$. Then $\{u,s_1,s_{i+1},\ldots ,s_k\} \sim _S S {\setminus } \{s_i\}$.

Proof

This is the same as Lemma 4.27 with u included in all $\mathcal L$ and $\mathcal R$ sets. The proof of Lemma 4.27 used Lemmas 4.24 and 4.18, so inclusion of u to the $\mathcal L$ and $\mathcal R$ sets is compatible with those lemmas (by Lemmas 4.34 and 4.31). The proof of Lemma 4.27 also used Lemma 4.17, and this does not immediately allow inclusion of u. We consider the two possibilities for $\ell$. If $\ell = 2$, then $\ell \le i$. In this case $s_{i+1}$, $\ldots$, $s_k$ do not have edges connecting to u, so the use of Lemma 4.17 to multiply by $\Delta _{\{s_{i+1},\ldots ,s_k\}}$ is valid even with u included in the $\mathcal L$ and $\mathcal R$ sets. If $\ell = 3$ then it is only if $i=2$ that there is an edge between $\{u\}$ and $\{s_{i+1},\ldots ,s_k\}$, namely between u and $s_3$. But in this case the required statement is $\{s_1,s_3,\ldots ,s_k\} \sim _{\{s_1,\ldots ,s_k\}} \{s_1,s_3,\ldots ,s_k\}$, which is true. $\square$

Proposition 4.38

Let G be an irreducible Artin–Tits group of finite type with generating set S having more than two elements. Let s, $t \in S$. Then $\{t\} \sim _S S {\setminus } \{s\}$.

Proof

If the Coxeter diagram of G is linear then this follows from Lemma 4.29. Now suppose that the Coxeter diagram of G is a tree having one vertex of valence three. We adopt the notation of Remark 4.30. We first consider the case where $s = s_i$ for some $1 \le i \le k$. Then

$$\begin{aligned} \{t\}&\sim _S \{s_\ell \}, \text { by 4.21,} \\&\sim _S \{u,s_\ell \}, \text { by Lemma 4.24,} \\&\sim _S \{u,s_1\}, \text { by Lemma 4.32,} \\&\sim _S \{u,s_1,s_k\}, \text { by Lemma 4.33,} \\&\sim _S \{u,s_1,s_{i+1},\ldots ,s_k\}, \text { by Lemma 4.36,} \\&\sim _S \{u,s_1,\ldots ,s_{i-1},s_{i+1},\ldots ,s_k\}, \text { by Lemma 4.37,} \\&= S {\setminus } \{s_i\}. \end{aligned}$$

Finally we treat the case where $s=u$. Consider the element $\Delta _{\{s_1,\ldots ,s_k\}}us_\ell$. Since $u \not \in \mathcal R(\Delta _{\{s_1,\ldots ,s_k\}})$, we know that $\Delta _{\{s_1,\ldots ,s_k\}}u \in P_{\text {red}}$. Next we note that $s_\ell \not \in \mathcal R(\Delta _{\{s_1,\ldots ,s_k\}}u)$, by the same argument used in the proof of Lemma 4.18. Therefore $\Delta _{\{s_1,\ldots ,s_k\}}us_\ell \in P_{\text {red}}$. The arguments used in the proof of Lemma 4.18 can be used to show that $\mathcal L(\Delta _{\{s_1,\ldots ,s_k\}}us_\ell ) = \{s_1,\ldots ,s_k\}$ and $\mathcal R(\Delta _{\{s_1,\ldots ,s_k\}}us_\ell ) = \{s_1,\ldots ,s_k\} {\setminus } \{s_{\ell -1},s_{\ell +1}\}$. Now we have $\{s_1,\ldots ,s_k\} {\setminus } \{s_{\ell -1},s_{\ell +1}\} \sim _{\{s_1,\ldots ,s_k\}} \{s_1,\ldots ,s_k\} {\setminus } \{s_\ell \}$ by Lemma 4.18, and $\{s_1,\ldots ,s_k\} {\setminus } \{s_\ell \} \sim _{\{s_1,\ldots ,s_k\}} \{s_1\}$ by Lemma 4.29. Combining these equivalences we have $S {\setminus } \{u\} = \{s_1,\ldots ,s_k\} \sim _S \{s_1,\ldots ,s_k\} {\setminus } \{s_{\ell -1},s_{\ell +1}\} \sim _S \{s_1,\ldots ,s_k\} {\setminus } \{s_\ell \} \sim _S \{s_1\} \sim _S \{t\}$, the last equivalence following from Lemma 4.21. $\square$

Theorem 4.39

Let G be an irreducible Artin–Tits group of finite type with at least three generators. Then $C^*_r(G \ltimes \tilde{\Omega }_{\infty })$ is simple and purely infinite.

Proof

Since $X_0$ is a compact-open transversal in $\tilde{\Omega }_{\infty }$ it is enough to consider the restriction to $X_0$. Thus we understand that the sets in the proof actually represent their intersections with $X_0$. We will consider a (nonempty) basic open set $U = Z(g) {\setminus } \bigcup _{i=1}^n Z(h_i)$ in $X_0$, as described in Remark 4.14 (thus g, $h_1$, $\ldots$, $h_n \in P {\setminus } \Delta P$). Let $y \in U$. Write the infinite normal form of y as $(y_1,y_2, \ldots )$. Choose $p > \max \{ \nu (g), \nu (h_1), \ldots , \nu (h_n) \}$. We claim that if $x \in X_0$ has infinite normal form beginning $(y_1, y_2, \ldots , y_p, \ldots )$ then $x \in U$. To prove this, consider such an element x. Since $g \in y$ we know that $g \prec y_1 \cdots y_r$ for some r. By Michel [40, Lemma 4.10] we have that $\nu (g) \le r$ and $g \prec y_1 \cdots y_{\nu (g)} \prec y_1 \cdots y_p$, and hence that $g \in x$, i.e. that $x \in Z(g)$. If for some i we have $x \in Z(h_i)$ then the same argument implies that $h_i \prec y_1 \cdots y_p$ and hence that $h_i \in y$, contradicting the fact that $y \not \in Z(h_i)$. Therefore $x \in Z(g) {\setminus } \bigcup _{i=1}^n Z(h_i) = U$. Next we claim that there is an element $\varepsilon \in P {\setminus } \Delta P$ such that $Z(\varepsilon ) \subseteq U$. To prove this, we first give a preliminary result. Consider an element $c \in P {\setminus } \Delta P$ with normal form $(c_1, \ldots , c_q)$, and suppose that for some $t \in S$ we have $\mathcal L(c_q) = S {\setminus } \{ t \}$. We claim that every element of Z(c) has infinite normal form $(c_1, \ldots , c_{q-1}, \ldots )$. To see this, let $w \in P$ be such that $cw \in P {\setminus } \Delta P$. By Michel [40, Lemma 4.6] the normal form of $c_q w$ is $(c_q w_1', w_1'' w_2', \ldots )$. Then $S \supsetneq \mathcal L(c_q w_1') \supseteq \mathcal L(c_q) = S {\setminus } \{ t \}$, and hence $\mathcal L(c_q w_1') = S {\setminus } \{ t \}$. Since $(c_1, \ldots , c_q)$ is a normal form we must have $\mathcal R(c_{q-1}) = S {\setminus } \{ t \}$, and hence the normal form of cw is $(c_1, \ldots , c_{q-1}, c_q w_1', \ldots )$. Let $x \in Z(c)$ with infinite normal form $(x_1, x_2, \ldots )$. Then for all large enough r we have $x_1 \cdots x_r \in cP {\setminus } \Delta P$, and hence $x_i = c_i$ for $i < q$. Now we return to the proof of the existence of the element $\varepsilon$. By Proposition 4.38 there is $e \in P$ with normal form $(e_1, \ldots , e_k)$ such that $\mathcal R(y_p) \supseteq \mathcal L(e_1)$ and $\mathcal R(e_k) = S {\setminus } \{ t \}$ for some choice of $t \in S$. Let $\varepsilon = y_1 \cdots y_p e \Delta _{S {\setminus } \{ t \}}$. Let $x \in Z(\varepsilon )$. By the previous claim, x has infinite normal form $(y_1, \ldots , y_p, e_1, \ldots , e_k, \ldots )$. By the first claim above we have $x \in U$. Thus $Z(\varepsilon ) \subseteq U$. We will write the normal form of $\varepsilon$ as $(\varepsilon _1, \ldots , \varepsilon _m)$. It follows almost immediately that the restriction to $X_0$ is minimal: if $x \in X_0$, then $\varepsilon x \in Z(\varepsilon ) \subseteq U$. Now we prove that the restriction to $X_0$ is locally contractive. We must show that U contains an open set V such that $\overline{V} \subseteq U$, and such that there is $b \in P$ with $b \overline{V} \subsetneq V$. Since S has cardinality at least three we may choose $u \in S$ such that $u \not = t$ and $\mathcal L(\varepsilon _1) \not = \{u\}$. Then $u \in \mathcal R(\varepsilon _m)$. By Proposition 4.38 there is $a \in P$ with normal form $(a_1,\ldots ,a_\ell )$ such that $\mathcal L(a_1) = \{u\}$ and $\mathcal R(a_\ell ) \supseteq \mathcal L(\varepsilon )$. Then elements of $\varepsilon u a Z(\varepsilon )$ have infinite normal forms $(\varepsilon _1,\ldots , \varepsilon _m, u, a_1, \ldots , a_\ell , \varepsilon _1, \ldots , \varepsilon _{m-1}, \ldots )$, while elements of $\varepsilon u^2 a Z(\varepsilon )$ have infinite normal forms $(\varepsilon _1,\ldots , \varepsilon _m, u, u, a_1, \ldots , a_\ell , \varepsilon _1, \ldots , \varepsilon _{m-1}, \ldots )$. This follows from the preliminary result above, applied to $c = \varepsilon u a \varepsilon$ and $c = \varepsilon u^2 a \varepsilon$. It follows that $\varepsilon u a Z(\varepsilon )$ and $\varepsilon u^2 a Z(\varepsilon )$ are disjoint subsets of $Z(\varepsilon )$. Therefore letting $b = \varepsilon u a$ we have that $b Z(\varepsilon ) \subsetneq Z(\varepsilon )$.

   Now we show that the restriction to $X_0$ is topologically principal. We must find an element $z \in U$ with trivial isotropy. Let $u \in \mathcal R(y_p)$, and let $t \in S$ with $m_{ut} > 2$. Let $(i_1,i_2,\ldots )$ be an aperiodic sequence in $\prod _1^\infty \{1,2\}$. In the following, for $s \in S$ we will write $(s)^i$ for the normal form $(s,s,\ldots ,s)$ with i terms. Let $z \in X_0$ have normal form $(y_1, \ldots , y_p, u, ut, (t)^{i_1}, tu, ut, (t)^{i_2}, tu, ut, \ldots )$. Then $z \in U$, by the first claim at the beginning of the proof of the Theorem. Assume $g \in G$ satisfies $gz = z$. Let z have normal form $(z_1, z_2, z_3, \dotsc )$. $z \in U_{g^{-1}}$ implies that $g = p q^{-1}$ and $q \in [z_1 \cdots z_l]$ for some l, i.e., there must exist $x \in P$ with $qx = z_1 \cdots z_l$. By replacing q by qx and p by px, we may assume without loss of generality that $q = z_1 \cdots z_l$. Then $q^{-1}z$ lies in $X_0$ and has normal form $(z_{l+1}, z_{l+2}, z_{l+3}, \dotsc ) = (z_{l+1}, \dotsc , z_n, ut, (t)^{i_j}, tu, ut, (t)^{i_{j+1}}, \dotsc )$. Let d be such that $d (q^{-1}z) \in X_0$. Let d have normal form $(d_1, \ldots , d_c)$. Let us consider the product $d_c z_{l+1} \cdots z_n (ut) t^{i_j} (tu)$. By Michel [40, Lemma 4.6] we can write $z_m = z_m' z_m''$ so that $d_c z_{l+1} \cdots z_n (ut) t^{i_j} (tu)$ has normal form either $(d_c z_{l+1}', z_{l+1}'' z_{l+2}', \ldots , z_{n-1}'' z_n', z_n'', ut, (t)^{i_j}, tu)$ or $(d_c z_{l+1}', z_{l+1}'' z_{l+2}', \ldots , z_{n-1}'' z_n', z_n'' u, t, (t)^{i_j}, tu)$ or $(d_c z_{l+1}', z_{l+1}'' z_{l+2}', \ldots , z_{n-1}'' z_n', z_n'' ut, (t)^{i_j}, tu)$. Similarly, the product $d_{c-1} d_c z_{l+1} \cdots z_n (ut) t^{i_j} (tu) (ut) t^{i_{j+1}} (tu)$ has normal form $( \cdots , ut, (t)^{i_{j+1}}, tu)$. Continuing we see that for all large enough s we have that $d z_{l+1} \cdots z_s$ has normal form $( \cdots , ut, (t)^{i_k}, tu, ut, (t)^{i_{k+1}}, tu, \ldots )$. So there exists a positive integer r with $(z_{r+1}, z_{r+2}, z_{r+3}, \ldots ) = (ut, (t)^{i_k}, tu, \ldots )$ and such that for all $s \ge r$, the normal form of $d z_{l+1} \cdots z_s$ is given by the normal form of $d z_{l+1} \cdots z_r$ followed by $z_{r+1}, z_{r+2}, z_{r+3}, \ldots , z_s$. So the normal form of dz is given by the normal form of $d z_{l+1} \cdots z_r$ followed by $ut, (t)^{i_k}, tu, ut, (t)^{i_{k+1}}, tu, \ldots$. Now apply this to $d=p$ and $d=q$. Then the normal form of $gz = p (q^{-1}z)$ is given by the normal form of $p z_{l+1} \cdots z_r$ followed by $ut, (t)^{i_k}, tu, ut, (t)^{i_{k+1}}, tu, \ldots$. Similarly, the normal form of $z = q (q^{-1}z)$ is given by the normal form of $q z_{l+1} \cdots z_r$ followed by $ut, (t)^{i_k}, tu, ut, (t)^{i_{k+1}}, tu, \ldots$. Since $gz = z$, these two normal forms must coincide. This — because of aperiodicity of $(i_h)$ — implies that the normal forms of $p z_{l+1} \cdots z_r$ and $q z_{l+1} \cdots z_r$ must coincide. Hence $p=q$, and thus $g=1$.

Therefore z has trivial isotropy. Now, simplicity follows from [46, Proposition II.4.6], and pure infiniteness from [2, Proposition 2.4]. $\square$

5 Exact Sequences in K-Theory

Let $P = \left\langle a,b \, \vert \, u=v \right\rangle ^+$ be a right LCM monoid as in Sect. 4, and assume that we are in case I or case II of Theorem 4.3. Let $G = \left\langle a,b \, \vert \, u=v \right\rangle$ be the group given by the same presentation as P. Suppose A is a C*-algebra and $\gamma : \, G \curvearrowright A$ is a G-action on A. The partial action $G \curvearrowright \Omega$ described in Sect. 2.2 induces a partial action $G \curvearrowright C(\Omega )$, which in turn gives rise to the diagonal partial action $G \curvearrowright C(\Omega ) \otimes A$. Recall that $\tilde{\Omega } = \Omega {\setminus } \left\{ \infty \right\}$ is G-invariant, so that we obtain a partial action $G \curvearrowright C_0(\tilde{\Omega }) \otimes A$ by restriction. As G is exact [28], we obtain an exact sequence

$$\begin{aligned} 0 \rightarrow (C_0(\tilde{\Omega }) \otimes A) \rtimes _r G \rightarrow (C(\Omega ) \otimes A) \rtimes _r G \rightarrow A \rtimes _r G \rightarrow 0. \end{aligned}$$

(8)

In case $P \subseteq G$ is Toeplitz, we know how to compute K-theory for $(C(\Omega ) \otimes A) \rtimes _r G$ because of [16, 17]. Therefore, our goal now is to deduce a six term exact sequence in K-theory which allows us to compute K-theory for $(C_0(\tilde{\Omega }) \otimes A) \rtimes _r G$. This will then enable us to determine the K-theory of $A \rtimes _r G$.

Let us explain our strategy. Recall that we defined $X = \left\{ \chi \in \Omega \text {: }\chi (wP) = 0 \right\}$, with $w \in P$ as in Lemma 4.1. As X is a compact open subspace of $\tilde{\Omega }$ meeting every G-orbit, the projection $\varvec{p} \mathrel {:=}(1_X \otimes 1)$ (where $1 = 1_{M(A)} \in M(A)$) is a full projection, so that $(C_0(\tilde{\Omega }) \otimes A) \rtimes _r G \sim _M \varvec{p} \left( (C_0(\tilde{\Omega }) \otimes A) \rtimes _r G \right) \varvec{p}$. Moreover, we have an exact sequence

$$\begin{aligned} 0\rightarrow & {} \varvec{p} \left( (C_0(P) \otimes A) \rtimes _r G \right) \varvec{p} \rightarrow \varvec{p} \left( (C_0(\tilde{\Omega }) \otimes A) \rtimes _r G \right) \varvec{p} \nonumber \\\rightarrow & {} \varvec{p} \left( (C_0(\tilde{\Omega }_{\infty }) \otimes A) \rtimes _r G \right) \varvec{p} \rightarrow 0. \end{aligned}$$

(9)

Note that we have an isomorphism

$$\begin{aligned} \mathcal K(\ell ^2(P \cap X)) \otimes A \cong \varvec{p} \left( (C_0(P) \otimes A) \rtimes _r G \right) \varvec{p}, \, e_{x,y} \otimes a \mapsto u_x (1_{\left\{ e \right\} } \otimes a) u_{y^{-1}}. \end{aligned}$$

(10)

Here $P \cap X = \left\{ x \in P \text {: }x \notin wP \right\}$, $e_{x,y}$ is the rank one operator corresponding to the basis vectors $\delta _x$ and $\delta _y$ in $\ell ^2(P \cap X)$, and $u_x, u_y$ are the canonical partial isometries implementing the partial action in the crossed product. Moreover, using the graph models from Sect. 4, it is possible to compute K-theory for $\varvec{p} \left( (C_0(\tilde{\Omega }_{\infty }) \otimes A) \rtimes _r G \right) \varvec{p}$. Hence the K-theoretic six term exact sequence provided by (9) already provides a way to compute K-theory for $\varvec{p} \left( (C_0(\tilde{\Omega }) \otimes A) \rtimes _r G \right) \varvec{p}$. However, since the C*-algebra we are interested in appears in the middle of the sequence, we would have to work out boundary maps, which can be complicated. Therefore, we present an alternative approach. The idea is to compare (9) with the canonical exact sequence for the twisted graph algebra of a graph model $\mathcal E$, where we twist $C^*(\mathcal E)$ by A using the G-action $\gamma$. In the following, let us make this precise, i.e., let us construct twisted graph algebras and their canonical extensions. Our construction generalizes the one in [14].

5.1 Graph algebras twisted by coefficients

Let $\mathcal E$ be a graph with set of vertices $\mathcal E^0$, set of edges $\mathcal E^1$ and range and source maps r, s. We assume that $\mathcal E$ is finite without sources because this will be the case in our applications. However, everything in this subsection also works for general graphs (with appropriate modifications).

Let F be the free group on the set of edges $\mathcal E^1$ of $\mathcal E$. Let $S_{\mathcal E}$ be the inverse semigroup attached to $\mathcal E$ as in [34, § 3.2], and $E_{\mathcal E}$ its semilattice of idempotents. Set $X_{\mathcal E} \mathrel {:=}\widehat{E_{\mathcal E}}$. $X_{\mathcal E}$ can be identified with the set of finite and infinite paths in $\mathcal E$. Let $W_{\mathcal E}$ be the subspace of finite paths and $Y_{\mathcal E}$ the subspace of infinite paths. Note that in our case, $Y_{\mathcal E} = \partial \mathcal E$. As explained in [34, § 3.2], there is a canonical partial action $F \curvearrowright X_{\mathcal E}$ such that $C(X_{\mathcal E}) \rtimes _r F$ is canonically isomorphic to the Toeplitz algebra of $\mathcal E$ and $C(Y_{\mathcal E}) \rtimes _r F$ is canonically isomorphic to the graph algebra of $\mathcal E$. Moreover, we have an exact sequence $0 \rightarrow C_0(W_{\mathcal E}) \rtimes _r F \rightarrow C(X_{\mathcal E}) \rtimes _r F \rightarrow C(Y_{\mathcal E}) \rtimes _r F \rightarrow 0$.

Now assume that A is a C*-algebra with an F-action $\gamma$. Then we can form the diagonal partial action $F \curvearrowright C(X_{\mathcal E}) \otimes A$. By exactness of F, we obtain an exact sequence

$$\begin{aligned} 0 \rightarrow (C_0(W_{\mathcal E}) \otimes A) \rtimes _r F \rightarrow (C(X_{\mathcal E}) \otimes A) \rtimes _r F \rightarrow (C(Y_{\mathcal E}) \otimes A) \rtimes _r F \rightarrow 0. \end{aligned}$$

(11)

Let $p_v$, $v \in \mathcal E^0$ and $s_e$, $e \in \mathcal E^1$, be the canonical generators (projections and partial isometries) of the Toeplitz algebra of $\mathcal E$, so that we may consider the elements $p_v$ and $s_e s_e^*$ in $C(X_{\mathcal E})$. A combination of [17, Corollary 3.14] and [36, § 6.1] (see also [35]) yields that the homomorphisms

$$\begin{aligned}&\bigoplus _{\mathcal E^0} A \rightarrow (C_0(W_{\mathcal E}) \otimes A) \rtimes _r F, \, (a_v)_v \mapsto \sum _v \Big ( p_v - \sum _{e \in r^{-1}(v)} s_e s_e^* \Big ) \otimes a_v,\end{aligned}$$

(12)

$$\begin{aligned}&\bigoplus _{\mathcal E^0} A \rightarrow (C(X_{\mathcal E}) \otimes A) \rtimes _r F, \, (a_v)_v \mapsto \sum _v p_v \otimes a_v \end{aligned}$$

(13)

induce isomorphisms in K-theory, and that these $K_*$-isomorphisms fit into a commutative diagram

(14)

where the lower horizontal map is the one in (11) and M is the $\mathcal E^0 \times \mathcal E^0$-matrix $M = (M_{w,v})_{w,v}$ with entries $M_{w,v} \in {\mathrm{End}}\,(K_*(A))$ given by $M_{w,v} = \sum _{e \in s^{-1}(w) \cap r^{-1}(v)} (\gamma _e)_*^{-1}$.

In order to compare the exact sequence (11) with (9), we need to establish a universal property for $(C(X_{\mathcal E}) \otimes A) \rtimes _r F$. Let us assume that A is unital and treat the non-unital case later. Let $\mathcal T_A$ be the universal C*-algebra which comes with a homomorphism $\iota : \, A \rightarrow \mathcal T_A$ such that $\mathcal T_A$ is generated by projections $p_v$, $v \in \mathcal E^0$, partial isometries $s_e$, $e \in \mathcal E^1$, and $\iota (A)$, subject to the relations that $p_v$, $v \in \mathcal E^0$, are pairwise orthogonal, $s_e^* s_e = p_{s(e)}$, $p_v \ge \sum _{e \in r^{-1}(v)} s_e s_e^*$ and $s_e \iota (a) = \iota (\gamma _e(a)) s_e$ for all $e \in \mathcal E^1$, $v \in \mathcal E^0$ and $a \in A$. This means that whenever we have a C*-algebra T, a homomorphism $i: \, A \rightarrow T$, projections $q_v$, $v \in \mathcal E^0$, and partial isometries $t_e$, $e \in \mathcal E^1$, such that $q_v$, $t_e$ and i(a) satisfy analogous relations as the generators of $\mathcal T_A$, then there is a (unique) homomorphism $\mathcal T_A \rightarrow T$ sending $p_v$ to $q_v$, $s_e$ to $t_e$ and $\iota (a)$ to i(a). Hence, by universal property of $\mathcal T_A$, there is a homomorphism

$$\begin{aligned} \mathcal T_A \rightarrow (C(X_{\mathcal E}) \otimes A) \rtimes _r F, \, p_v \mapsto 1_v \otimes 1_A \in C(X_{\mathcal E}) \otimes A, \, s_e \mapsto u_e, \, \iota (a) \mapsto 1 \otimes a. \end{aligned}$$

(15)

Here $1_v$ is the characteristic function of the subspace of $X_{\mathcal E}$ of all finite and infinite paths with range v. We see that $\iota : \, A \rightarrow \mathcal T_A$ must be injective, because $A \rightarrow (C(X_{\mathcal E}) \otimes A) \rtimes _r F, a \mapsto 1 \otimes a$ is. So we will from now on identify A with $\iota (A)$ and write a instead of $\iota (a)$. Our goal is to show that the homomorphism in (15) is an isomorphism.

For a finite path $\mu \in \mathcal E^*$, where $\mu = \mu _1 \cdots \mu _m$ with $\mu _i \in \mathcal E^1$, let $s_{\mu } = s_{\mu _1} \cdots s_{\mu _m}$. First of all, it is easy to see that $\mathcal T_A = \overline{\mathrm{span}}\big ( \left\{ s_{\mu } s_{\nu }^* a \text {: }a \in A, \, \mu , \nu \in \mathcal E^*, \, s(\mu ) = s(\nu ) \right\} \big )$. Moreover, by universal property of $\mathcal T_A$, for every $z \in \mathbb {T}$ there is an isomorphism $\delta _z: \, \mathcal T_A \cong \mathcal T_A$ given by $\delta _z(a) = a$, $\delta _z(p_v) = p_v$ and $\delta _z(s_e) = z s_e$. The following is easy to see (as in the case of ordinary graph algebras without coefficients):

Lemma 5.1

$\theta (x) \mathrel {:=}\int _{\mathbb {T}} \delta _z(x) dz$ defines a faithful conditional expectation $\mathcal T_A \twoheadrightarrow \mathcal T_A^{\delta }$ determined by $\theta (s_{\mu } s_{\nu }^* a) = \delta _{l(\mu ), \, l(\nu )} s_{\mu } s_{\nu }^* a$.

Now let $\mathcal T$ be the classical Toeplitz algebra of $\mathcal E$, so that $\mathcal T$ is the universal C*-algebra generated by projections $p_v$, $v \in \mathcal E^0$, and partial isometries $s_e$, $e \in \mathcal E^1$, satisfying the same relations as the generators of $\mathcal T_A$. Again, we have a canonical $\mathbb {T}$-action $\delta$ on $\mathcal T$, with fix point algebra $\mathcal T^{\delta }$.

Lemma 5.2

There is an isomorphism $\mathcal T^{\delta } \otimes A \rightarrow \mathcal T_A^{\delta }$ sending $s_{\mu } s_{\nu }^* \otimes a$ to $s_{\mu } a s_{\nu }^*$.

Proof

Let

$$\begin{aligned} \mathcal T_l^{\delta }&= \overline{\mathrm{span}}\left\{ s_{\mu } s_{\nu }^* \text {: }s(\mu ) = s(\nu ), \, l(\mu ) = l(\nu ) \le l \right\} ,\\ \mathcal T_{A,l}^{\delta }&= \overline{\mathrm{span}}\left\{ s_{\mu } s_{\nu }^* a \text {: }a \in A, \, s(\mu ) = s(\nu ), \, l(\mu ) = l(\nu ) \le l \right\} . \end{aligned}$$

Then $\mathcal T^{\delta } \otimes A \cong \varinjlim _l \mathcal T_l^{\delta } \otimes A$ and $\mathcal T_A^{\delta } \cong \varinjlim _l \mathcal T_{A,l}^{\delta }$. So it suffices to show that for every l, we have an isomorphism $\mathcal T_l^{\delta } \otimes A \rightarrow \mathcal T_{A,l}^{\delta }, \, s_{\mu } s_{\nu }^* \otimes a \mapsto s_{\mu } a s_{\nu }^*$. For fixed l, define $\epsilon _{\mu , \nu } = s_{\mu } s_{\nu }^* - \sum _{e \in r^{-1}(s(\mu ))} s_{\mu e} s_{\nu e}^*$ if $l(\mu ) = l(\nu ) \le l-1$ and $\epsilon _{\mu , \nu } = s_{\mu } s_{\nu }^*$ if $l(\mu ) = l(\nu ) = l$. Then we have $\epsilon _{\kappa , \lambda }^* = \epsilon _{\lambda , \kappa }$ and $\epsilon _{\kappa , \lambda } \epsilon _{\mu , \nu } = \delta _{\lambda , \mu } \epsilon _{\kappa ,\nu }$. Thus $\mathcal T_l^{\delta } \otimes A$ is the universal C*-algebra generated by $\epsilon _{\mu ,\nu } \otimes a$ for $\mu , \nu \in \mathcal E^*$ with $s(\mu ) = s(\nu )$ and $l(\mu ) = l(\nu ) \le l$, and for $a \in A$, subject to the relations $(\epsilon _{\kappa , \lambda } \otimes a)^* = \epsilon _{\lambda , \kappa } \otimes a^*$ and $(\epsilon _{\kappa , \lambda } \otimes a) (\epsilon _{\mu , \nu } \otimes b) = \delta _{\lambda , \mu } \epsilon _{\kappa ,\nu } \otimes ab$. This shows existence of the homomorphism $\mathcal T_l^{\delta } \otimes A \rightarrow \mathcal T_{A,l}^{\delta }, \, s_{\mu } s_{\nu }^* \otimes a \mapsto s_{\mu } a s_{\nu }^*$. It is clearly surjective. To prove injectivity, it suffices to show that $A \rightarrow \mathcal T_{A,l}^{\delta }, \, a \mapsto s_{\mu } a s_{\nu }^* - \sum _{e \in r^{-1}(s(\mu ))} s_{\mu e} a s_{\nu e}^*$ (for $l(\mu ) = l(\nu ) \le l-1$) and $A \rightarrow \mathcal T_{A,l}^{\delta }, \, a \mapsto s_{\mu } a s_{\nu }^*$ (for $l(\mu ) = l(\nu ) = l$) are injective. But this is easy to see by considering their compositions $A \rightarrow \mathcal T_{A,l}^{\delta } \rightarrow \mathcal T_A^{\delta } \rightarrow \mathcal T_A \rightarrow (C(X_{\mathcal E}) \otimes A) \rtimes _r F$. $\square$

Corollary 5.3

The homomorphism in (15) is an isomorphism.

Proof

Consider the composition $\Theta : \, \mathcal T_A \overset{\theta }{\longrightarrow } \mathcal T_A^{\delta } \cong \mathcal T^{\delta } \otimes A \rightarrow C(X_{\mathcal E}) \otimes A$, where the last map is the tensor product of the canonical faithful conditional expectation $\mathcal T^{\delta } \twoheadrightarrow C(X_{\mathcal E})$ with $\mathrm{id}_A$. $\Theta$ is a faithful conditional expectation determined by $s_{\mu } a s_{\nu }^* \mapsto \delta _{\mu ,\nu } 1_{\mu } \otimes a$, where $1_{\mu }$ is the characteristic function of the subspace of $X_{\mathcal E}$ of all finite and infinite paths starting with $\mu$. It is now easy to see that we have a commutative diagram

where the right vertical map is the canonical faithful conditional expectation and the upper horizontal map is the homomorphism in (15). It follows that the map in (15), which is clearly surjective, is also injective. $\square$

5.2 Comparing exact sequences in K-theory

Now we return to the setting described at the beginning of Sect. 5. Let $\mathcal E$ be a graph model for $(G \ltimes \tilde{\Omega }_{\infty }) \, \vert \, Y$, where $Y = \Omega _{\infty } \cap X = \tilde{\Omega }_{\infty } \cap X$. The G-action $\gamma$ induces an action of the free group F on $\mathcal E^1$ on A by letting $e \in \mathcal E^1$ act via $\gamma _{\varvec{\sigma }(e)}$.

First let A be unital. By universal property of $(C(X_{\mathcal E}) \otimes A) \rtimes _r F$, as established in Sect. 5.1, we have a homomorphism

$$\begin{aligned} (C(X_{\mathcal E}) \otimes A) \rtimes _r F \rightarrow \varvec{p} \left( (C_0(\tilde{\Omega }) \otimes A) \rtimes _r G \right) \varvec{p}, \, p_v \mapsto 1_X 1_{vP} \otimes 1_A, \, u_e \mapsto \varvec{p} u_{\varvec{\sigma }(e)} (1_{s(e)P} \otimes 1_A) \varvec{p}, \, a \mapsto 1_X \otimes a. \end{aligned}$$

Note that in general, this homomorphism is neither injective nor surjective. By our choice of $\mathcal E$, this homomorphism fits into a commutative diagram

(16)

where the third vertical map is an isomorphism because $\mathcal E$ is a graph model for $(G \ltimes \tilde{\Omega }_{\infty }) \, \vert \, Y$.

For non-unital A, just apply our constructions above to the unitalization $A \tilde{\ }$ and then restrict all the maps to the crossed products with A in place of $A \tilde{\ }$.

The isomorphism (10) implies that the homomorphism $A \rightarrow \varvec{p} \left( (C_0(P) \otimes A) \rtimes _r G \right) \varvec{p}, \, a \mapsto 1_{\left\{ e \right\} } \otimes a$ induces an $K_*$-isomorphism, which together with the $K_*$-isomorphism induced by (12) fit into a commutative diagram

(17)

where the right vertical map is induced by the first vertical map in (16), and $\phi = \sum _{v \in \mathcal E^0} \phi _v$ with

$$\begin{aligned} \phi _v = \sum _{x \in \left( vP \, {\setminus } \, \bigcup _{e \in r^{-1}(v)} \varvec{\sigma }(e) s(e) P \right) \cap X} (\gamma _x)_*^{-1}. \end{aligned}$$

Let us now assume that our graph model $\mathcal E$ is such that the map $\phi$ in (17) is surjective. This is the only requirement we need on our graph models. In Sect. 5.3, we will construct concrete graph models with this property.

Here is a general lemma which we will apply in our special situation:

Lemma 5.4

Suppose we have the following commutative diagram of abelian groups with exact rows:

(18)

Then we have an exact sequence

(19)

This lemma is most likely well-known, but for the sake of completeness, we include a proof.

Proof

Since $\phi _{i+1} \circ \partial _i \circ \psi _i^{-1} \circ q_i = \epsilon _i \circ \psi _i \circ \psi _i^{-1} \circ q_i = \epsilon _i \circ q_i = 0$, we have $\mathrm{im\,}(\partial _i \circ \psi _i^{-1} \circ q_i) \subseteq \mathrm{ker}\,(\phi _{i+1})$. Moreover, composition of consecutive homomorphisms gives zero since $\partial _i \circ \psi _i^{-1} \circ q_i \circ \pi _i = \partial _i \circ \psi _i^{-1} \circ \psi _i \circ p_i = \partial _i \circ p_i = 0$.

Let us verify exactness at $G_i$: Take $z \in G_i$ with $\pi _i(z) = 0$. Then $\psi _i(p_i(z)) = q_i(\pi _i(z)) = 0$, so that $p_i(z) = 0$. So there exists $y \in \check{G}_i$ with $j_i(y) = z$. Since $k_i(\phi _i(y)) = \pi _i(j_i(y)) = 0$, there is $x \in \bar{H}_{i-1}$ with $\epsilon _{i-1}(x) = \phi _i(y)$. Set $w \mathrel {:=}\psi _{i-1}^{-1}(x)$ and $v \mathrel {:=}y - \partial _{i-1}(w)$. Then $\phi _i(v) = \phi _i(y) - \phi _i(\partial _{i-1}(w)) = \phi _i(y) - \epsilon _{i-1}(\psi _{i-1}(w)) = \phi _i(y) - \epsilon _{i-1}(x) = 0$ and $j_i(v) = j_i(y) = z$.

Now we show exactness at $H_i$: Let $z \in H_i$ satisfy $\partial _i(\psi _i^{-1}(q_i(z))) = 0$. Then there exists $y \in G_i$ with $p_i(y) = \psi _i^{-1}(q_i(z))$. Since $q_i(\pi _i(y)) = \psi _i(p_i(y)) = q_i(z)$, there is $x \in \check{H}_i$ with $k_i(x) = z - \pi _i(y)$. As $\phi _i$ is surjective, there is $w \in \check{G}_i$ with $\phi _i(w) = x$. Let $v \mathrel {:=}y + j_i(w)$. Then $\pi _i(v) = \pi _i(y) + \pi _i(j_i(w)) = \pi _i(y) + k_i(\phi _i(w)) = \pi _i(y) + k_i(x) = z$.

Finally, for exactness at $\mathrm{ker}\,(\phi _{i+1})$, let $z \in \mathrm{ker}\,(\phi _{i+1})$ satisfy $j_{i+1}(z) = 0$. Then there is $y \in \bar{G}_i$ with $\partial _i(y) = z$. Since $\epsilon _i(\psi _i(y)) = \phi _{i+1}(\partial _i(y)) = \phi _{i+1}(z) = 0$, there is $x \in H_i$ with $q_i(x) = \psi _i(y)$. Then $\partial _i(\psi _i^{-1}(q_i(x))) = \partial _i(\psi _i^{-1}(\psi _i(y))) = \partial _i(y) = z$. $\square$

Let us now apply Lemma 5.4 to the diagram induced by (16) in K-theory, where $\mathcal E$ is a graph model for $(G \ltimes \tilde{\Omega }_{\infty }) \, \vert \, Y$ such that the map $\phi$ in (17) is surjective. Let us write $I \mathrel {:=}\varvec{p} \left( (C_0(\tilde{\Omega }) \otimes A) \rtimes _r G \right) \varvec{p}$. Plugging (14) and (17) into the K-theory diagram for (16), we obtain the following commutative diagram with exact rows

(20)

Applying Lemma 5.4, we obtain an exact sequence

(21)

Furthermore, write $J \mathrel {:=}(C_0(\tilde{\Omega }) \otimes A) \rtimes _r G$. Since $\varvec{p}$ is a full projection in J, the canonical inclusion $I \hookrightarrow J$ induces a $K_*$-isomorphism. Let us assume that $P \subseteq G$ is Toeplitz, which in our case amounts to saying that $P \subseteq G$ is quasi-lattice ordered as in Remark 3.6. Then [17, Corollary 3.14] implies that the homomorphism $A \rightarrow (C(\Omega ) \otimes A) \rtimes _r G, \, a \mapsto 1 \otimes a$ induces an isomorphism in K-theory (because G satisfies the Baum–Connes conjecture with coefficients by [3]). Hence, applying K-theory to (8) and using our observations above, we obtain a second exact sequence

(22)

where $K_i(A) \rightarrow K_i(A \rtimes _r G)$ is induced by the canonical map and $\iota _i \circ \pi _i$ is given by

$$\begin{aligned} \iota _i \circ \pi _i = \sum _{v \in \mathcal E^0} \left( (\gamma _v)_*^{-1} - (\gamma _{v \vee w})_*^{-1} \right) : \, \bigoplus _{v \in \mathcal E^0} K_i(A) \rightarrow K_i(A), \end{aligned}$$

(23)

where $v \vee w \in P$ is the least common multiple of v and w determined by $(v \vee w) P = vP \cap wP$.

5.3 K-theory exact sequences for Artin–Tits groups of dihedral type and torus knot groups

Let us now consider two classes of monoids and groups which already appeared in Example 4.5: Let $P = \left\langle a,b \, \vert \, u=v \right\rangle ^+$ and $G = \left\langle a,b \, \vert \, u=v \right\rangle$, where $u = aba \cdots$ and $v = bab \cdots$ with $\ell ^*(u) = \ell ^*(v)$, or $u = a^p$ and $v = b^q$ for $p, q \ge 2$. In the first case, P and G are called Artin–Tits monoids and Artin–Tits groups of dihedral type $I_2(m)$ with $m = \ell ^*(u) = \ell ^*(v)$, and in the second case, P and G are called torus knot monoids or torus knot groups of type (p, q). We refer to the first case as the dihedral Artin–Tits case of type $I_2(m)$ and to the second case as the torus knot case of type (p, q). In both of these cases, our goal is to provide concrete graph models for $(G \ltimes \tilde{\Omega }_{\infty }) \, \vert \, Y$ such that $\phi$ in (17) is surjective, and then to work out the exact sequences (21) and (22) explicitly and further simplify them. Note that by Cuntz et al. [18, § 5.8], $P \subseteq G$ is Toeplitz because P is right reversible (not only left reversible). We start by constructing graph models.

Lemma 5.5

In the dihedral Artin–Tits case of type $I_2(m)$, with m even, construct a graph by setting $\mathcal E^0$ as the following collection of finite words in $\left\{ a,b \right\} ^*$ of length at most $m-1$:

$$\begin{aligned}&aa, abaa, ababaa, \dotsc , \underbrace{ab \cdots ba}_{m-1}, \underbrace{ab \cdots bb}_{m-1}, \dotsc , ababb, abb, baa, babaa, \dotsc ,\nonumber \\&\quad \underbrace{ba \cdots baa}_{m-1}, \underbrace{ba \cdots bab}_{m-1}, \dotsc , babb, bb. \end{aligned}$$

(24)

Let

$$\begin{aligned} \mathcal E^1 =&\left\{ (y,x) \text {: }y, x \in \mathcal E^0; \; \ell ^*(y) = \ell ^*(x) + 1; \; \exists \, \sigma \in \left\{ a,b \right\} : \, y \equiv \sigma x \right\} \\ \cup&\left\{ (y,x) \text {: }y, x \in \mathcal E^0; \; \ell ^*(y) = \ell ^*(x); \; \exists \, \sigma , \tau \in \left\{ a,b \right\} : \, y \tau \equiv \sigma x \text { is } u,v \text {-free} \right\} \\ \cup&\left\{ (aa,x) \text {: }x \in \mathcal E^0; \; x \text { starts with } ab \right\} \cup \left\{ (bb,x) \text {: }x \in \mathcal E^0; \; x \text { starts with } ba \right\} . \end{aligned}$$

Define range and source maps by $r(y,x) = y$ and $s(y,x) = x$, and set $\varvec{\sigma }(y,x)$ as the first letter of y. Then $(\mathcal E,\varvec{\sigma })$ is a graph model for $G \curvearrowright Y$. In the dihedral Artin–Tits case of type $I_2(m)$, with m odd, construct a graph by setting $\mathcal E^0$ as the following collection of finite words in $\left\{ a,b \right\} ^*$ of length at most $m-1$:

$$\begin{aligned}&aa, abaa, ababaa, \dotsc , \underbrace{ab \cdots baa}_{m-1}, \underbrace{ab \cdots ab}_{m-1}, \dotsc , ababb, abb, baa, babaa, \dotsc ,\nonumber \\&\quad \underbrace{ba \cdots ba}_{m-1}, \underbrace{ba \cdots bb}_{m-1}, \dotsc , babb, bb. \end{aligned}$$

(25)

Define $\mathcal E^1$, r, s and $\varvec{\sigma }$ as in the dihedral Artin–Tits case of type $I_2(m)$, with m even. Then $(\mathcal E,\varvec{\sigma })$ is a graph model for $G \curvearrowright Y$. In the torus knot case of type (p, q), construct a graph by setting $\mathcal E^0$ as the following collection of finite words in $\left\{ a,b \right\} ^*$:

$$\begin{aligned} ab, a^2b, \dotsc , a^{p-2}b, a^{p-1}, ba, b^2a, \dotsc , b^{q-1}a. \end{aligned}$$

(26)

Let

$$\begin{aligned} \mathcal E^1 =&\left\{ (y,x) \text {: }y, x \in \mathcal E^0; \; \ell ^*(y) = \ell ^*(x) + 1; \; \exists \, \sigma \in \left\{ a,b \right\} : \, y \equiv \sigma x \right\} \\ \cup&\left\{ (y,x) \text {: }y, x \in \mathcal E^0; \; \ell ^*(y) = \ell ^*(x); \; \exists \, \sigma , \tau \in \left\{ a,b \right\} : \, y \tau \equiv \sigma x \text { is } u,v \text {-free} \right\} \\ \cup&\left\{ (ab,x) \text {: }x \in \mathcal E^0; \; x \text { starts with } b \right\} \cup \left\{ (ba,x) \text {: }x \in \mathcal E^0; \; x \text { starts with } a \right\} . \end{aligned}$$

Define r, s and $\varvec{\sigma }$ as in the dihedral Artin–Tits case of type $I_2(m)$, with m even. Then $(\mathcal E,\varvec{\sigma })$ is a graph model for $G \curvearrowright Y$.

Proof

We first treat the dihedral Artin–Tits case. As in case I of Theorem 4.3, we have a bijection

$$\begin{aligned} \left\{ z \in \left\{ a,b \right\} ^{\infty } \text {: }z \text { is } u,v \text {-free} \right\} , \, z \mapsto \chi _z. \end{aligned}$$

Moreover, the inverse of $\mathcal E^{\infty } \rightarrow \left\{ z \in \left\{ a,b \right\} ^{\infty } \text {: }z \text { is } u,v \text {-free} \right\} , \, \mu \mapsto \varvec{\sigma }(\mu )$ is given as follows: Every u, v-free infinite word $z \in \left\{ a,b \right\} ^{\infty }$ is of the form $z = z_1 \sigma _1^{e_1} z_2 \sigma _2^{e_2} \cdots$ where $z_i \in \left\{ a,b \right\} ^*$ are u, v-free words ending on aa or bb and containing no other aa or bb as subwords, $\sigma _i$ is the last letter of $z_i$ and differs from the first letter of $z_{i+1}$, and $e_i \in \mathbb {Z}_{\ge 0} \cup \left\{ \infty \right\}$ ($e_i = \infty$ means that $z = \cdots z_i \sigma _i \sigma _i \sigma _i \cdots$). For each $z_i$, write $z_i = \tau _1 \cdots \tau _l$ with $\tau _k \in \left\{ a,b \right\}$, and for $1 \le j \le l-1$, set $x_j \mathrel {:=}\tau _j \cdots \tau _l$ if $l-j \le m-2$; if $l-j > m-2$, then we must have $l=m$ and $j=1$, in this case set $x_1 = \tau _1 \cdots \tau _{l-1}$. Now define $\varvec{\mu }(z_i) = (x_1,x_2) \cdots (x_{l-2},x_{l-1})$. Finally, set

$$\begin{aligned} \varvec{\mu }(z)&= \varvec{\mu }(z_1) (\sigma _1 \sigma _1, \sigma _1 \sigma _1)^{e_1} (\sigma _1 \sigma _1, (\sigma _1 z_2)_{[1,m-1]}) ((\sigma _1 z_2)_{[1,m-1]}, z_2) \cdot \\&\quad \cdot \varvec{\mu }(z_2) (\sigma _2 \sigma _2, \sigma _2 \sigma _2)^{e_2} (\sigma _2 \sigma _2, (\sigma _2 z_3)_{[1,m-1]}) ((\sigma _2 z_3)_{[1,m-1]}, z_3) \cdots . \end{aligned}$$

This defines a map $\varvec{\mu }: \, \left\{ z \in \left\{ a,b \right\} ^{\infty } \text {: }z \text { is } u,v \text {-free} \right\} \rightarrow \mathcal E^{\infty }$ which is the desired inverse. The rest of the argument proceeds exactly as in the proof of case I of Theorem 4.3.

Next, we treat the torus knot case of type (p, q). As in case I of Theorem 4.3, we have a bijection

$$\begin{aligned} \left\{ z \in \left\{ a,b \right\} ^{\infty } \text {: }z \text { is } u,v \text {-free} \right\} , \, z \mapsto \chi _z. \end{aligned}$$

Moreover, the inverse of $\mathcal E^{\infty } \rightarrow \left\{ z \in \left\{ a,b \right\} ^{\infty } \text {: }z \text { is } u,v \text {-free} \right\} , \, \mu \mapsto \varvec{\sigma }(\mu )$ is given as follows: Every u, v-free infinite word $z \in \left\{ a,b \right\} ^{\infty }$ is of the form $z = z_1 z_2 \cdots$ where $z_i$ are powers of $\sigma _i \in \left\{ a,b \right\}$, i.e., $z_i \in \left\{ \varepsilon , a^1, \dotsc , a^{p-1}, b^1, \dotsc , b^{q-1} \right\}$, and we must have $\sigma _i \ne \sigma _{i+1}$. For each $z_i$, write $z_i \sigma _{i+1} = \tau _1 \cdots \tau _l$ with $\tau _k \in \left\{ a,b \right\}$. If $z_i \sigma _{i+1} \ne a^{p-1}b$, let $x_j \mathrel {:=}\tau _j \cdots \tau _l$ for all $1 \le j \le l-1$. If $z_i \sigma _{i+1} = a^{p-1}b$, let $x_1 = a^{p-1}$ and $x_j \mathrel {:=}\tau _j \cdots \tau _l$ for all $2 \le j \le l-1$. Now define $\mu _i = (x_1,x_2) \cdots (x_{l-2},x_{l-1})$. Finally, set $\varvec{\mu }(w) = \mu _1 \, (\sigma _1 \sigma _2, r(\mu _2)) \, \mu _2 \, (\sigma _2 \sigma _3, r(\mu _3)) \cdots$. This defines a map $\varvec{\mu }: \, \left\{ z \in \left\{ a,b \right\} ^{\infty } \text {: }z \text { is } u,v \text {-free} \right\} \rightarrow \mathcal E^{\infty }$ which is the desired inverse. The rest of the argument proceeds exactly as in the proof of case I of Theorem 4.3. $\square$

Now let $\gamma : \, G \curvearrowright A$ be an action of G on a C*-algebra A, and we write $\alpha _i \mathrel {:=}(\gamma _a)_*^{-1}$ and $\beta _i \mathrel {:=}(\gamma _b)_*^{-1}$ for the automorphisms $K_i(A) \cong K_i(A)$ induced by $\gamma _a^{-1}$ and $\gamma _b^{-1}$ in K-theory.

Theorem 5.6

For the concrete graph models from Lemma 5.5, $\phi$ in (17) is surjective and (21) simplifies to

(27)

where

$$\begin{aligned} \tilde{j}_i = \begin{pmatrix} \mathrm{id}- \underbrace{\beta _i \alpha _i \cdots \beta _i \alpha _i}_{m} \\ \mathrm{id}+ \beta _i \alpha _i + \cdots + (\beta _i \alpha _i)^{\frac{m-2}{2}} - \beta _i \left( \mathrm{id}+ \alpha _i \beta _i + \cdots + (\alpha _i \beta _i)^{\frac{m-2}{2}} \right) \end{pmatrix} \end{aligned}$$

(28)

in the dihedral Artin–Tits case of type $I_2(m)$ with m even,

$$\begin{aligned} \tilde{j}_i = \begin{pmatrix} \mathrm{id}+ \underbrace{\beta _i \alpha _i \cdots \alpha _i \beta _i}_{m} \\ \mathrm{id}+ \beta _i \alpha _i + \cdots + (\beta _i \alpha _i)^{\frac{m-1}{2}} - \beta _i \left( \mathrm{id}+ \alpha _i \beta _i + \cdots + (\alpha _i \beta _i)^{\frac{m-3}{2}} \right) \end{pmatrix} \end{aligned}$$

(29)

in the dihedral Artin–Tits case of type $I_2(m)$ with m odd, and

$$\begin{aligned} \tilde{j}_i = \begin{pmatrix} \mathrm{id}+ \alpha _i + \cdots + \alpha _i^{p-1} \\ \mathrm{id}+ \beta _i + \cdots + \beta _i^{q-1} \end{pmatrix} \end{aligned}$$

(30)

the torus knot case of type (p, q).

In addition, we have the exact sequence

Note the asymmetry between a and b in (26), which is necessary to make sure that $\phi$ in (17) is surjective. In the following, we summarize the main steps in the computations leading to Theorem 5.6. We point out that everything is constructive, so that it is possible to keep track of all the identifications made along the way.

Proof

In the following, we write $\alpha = \alpha _i$ and $\beta = \beta _i$ to simplify notation. We have (22) because in our cases, P is right reversible (not only left reversible), so that $P \subseteq G$ is Toeplitz by Cuntz et al. [18, § 5.8].

We first treat the dihedral Artin–Tits case. With respect to the ordering of $\mathcal E^0$ as in (24) and (25), the original homomorphism $j_i$ in (20) is given by

(31)

where the row is at the position corresponding to the finite word $\underbrace{ab \cdots ba}_{m-1}$ in (24) if m is even and to $\underbrace{ab \cdots ab}_{m-1}$ in (25) if m is odd, and the row is at the position corresponding to the finite word $\underbrace{ba \cdots bab}_{m-1}$ in (24) if m is even and to $\underbrace{ba \cdots ba}_{m-1}$ in (25) if m is odd. By performing elementary row operations, which correspond to post-composition with isomorphisms, $j_i$ is transformed to

(32)

where we write 1 for $\mathrm{id}$ and

$$\begin{aligned} \varvec{w}= & {} 1 - \alpha - \beta \alpha ^2 - \cdots - \underbrace{\beta \alpha \cdots \beta \alpha ^2}_{m-1}, \ \varvec{x} = - \beta ^2 - \beta \alpha \beta ^2 - \cdots - \underbrace{\beta \alpha \cdots \beta \alpha \beta ^2}_{m-2},\\ \varvec{y}= & {} - \alpha ^2 - \alpha \beta \alpha ^2 - \cdots - \underbrace{\alpha \beta \cdots \alpha \beta \alpha ^2}_{m-2}, \ \varvec{z} = 1 - \beta - \alpha \beta ^2 - \cdots - \underbrace{\alpha \beta \cdots \alpha \beta ^2}_{m-1} \quad \mathrm{if} \ m \ \text {is even, and}\\ \varvec{w}= & {} 1 - \alpha - \beta \alpha ^2 - \cdots - \underbrace{\beta \alpha \cdots \beta \alpha ^2}_{m-2}, \ \varvec{x} = - \beta ^2 - \beta \alpha \beta ^2 - \cdots - \underbrace{\beta \alpha \cdots \beta \alpha \beta ^2}_{m-1},\\ \varvec{y}= & {} - \alpha ^2 - \alpha \beta \alpha ^2 - \cdots - \underbrace{\alpha \beta \cdots \alpha \beta \alpha ^2}_{m-1}, \ \varvec{z} = 1 - \beta - \alpha \beta ^2 - \cdots - \underbrace{\alpha \beta \cdots \alpha \beta ^2}_{m-2} \quad \mathrm{if} \ m \ \text {is odd}. \end{aligned}$$

The map $\phi _i$ in (20) is given by $\phi _i = (\varvec{\chi }, 0, \dotsc , 0, \bar{\omega }, 0, \dotsc , 0, \omega , 0, \dotsc , 0, \varvec{\psi })$, where

$$\begin{aligned} \varvec{\chi }= & {} \alpha ^2 + \beta \alpha ^2 + \alpha \beta \alpha ^2 + \cdots + \underbrace{\cdots \beta \alpha \beta \alpha ^2}_{m-1}, \ \varvec{\psi } = \beta ^2 + \alpha \beta ^2 + \beta \alpha \beta ^2 + \cdots + \underbrace{\cdots \alpha \beta \alpha \beta ^2}_{m-1},\\ \bar{\omega }= & {} \underbrace{\cdots \beta \alpha \beta \alpha }_{m-1} \ \text {is at the position corresponding to the finite word} \ \underbrace{ab \cdots ba}_{m-1} \ \mathrm{in} \ (24) \ \\&\mathrm{if} \ m \ \text {is even} \ \text {and to} \ \underbrace{ab \cdots ab}_{m-1} \ \mathrm{in} \ (25) \ \mathrm{if} \ m \ \text {is odd, and} \\ \omega= & {} \underbrace{\cdots \alpha \beta \alpha \beta }_{m-1} \ \text {is at the position corresponding to the finite word} \ \underbrace{ba \cdots bab}_{m-1}\ \mathrm{in} \ (24)\ \\&\mathrm{if} \ m \ \text {is even} \ \text {and to} \ \underbrace{ba \cdots ba}_{m-1} \ \mathrm{in} \ (25) \ \mathrm{if} \ m \ \text {is odd}. \end{aligned}$$

Now the following map $\bigoplus _{\mathcal E^0 {\setminus } \left\{ v \right\} } K_i(A) \rightarrow \bigoplus _{\mathcal E^0} K_i(A)$ induces an isomorphism $\bigoplus _{\mathcal E^0 {\setminus } \left\{ v \right\} } K_i(A) \cong \mathrm{ker}\,(\phi _i)$, where v is the finite word $\underbrace{ba \cdots bab}_{m-1}$ in (24) if m is even and to $\underbrace{ba \cdots ba}_{m-1}$ in (25) if m is odd:

where the double vertical line indicates the position of v. Pre-composing $j'_i$ by this isomorphism yields the homomorphism $\tilde{j'_i}: \, \bigoplus _{\mathcal E^0 {\setminus } \left\{ v \right\} } K_i(A) \cong \mathrm{ker}\,(\phi _i) \rightarrow \bigoplus _{\mathcal E^0} K_i(A)$ given by

Since we have many columns with only one non-zero entry given by 1, we can first perform elementary column operations and then permutations so that $\tilde{j'_i}$ is transformed to a matrix whose upper left corner consists of the identity matrix and the lower right corner is of the form

$$\begin{aligned} \left( \begin{array}{ccc} \varvec{w} &{}\quad 0 &{}\quad \varvec{x} \\ -\alpha &{}\quad 1 &{}\quad 0 \\ - \omega ^{-1} \varvec{\chi } &{}\quad - \omega ^{-1} \bar{\omega } &{}\quad - \beta - \omega ^{-1} \varvec{\psi } \\ \varvec{y} &{}\quad 0 &{}\quad \varvec{z} \end{array} \right) . \end{aligned}$$

Let us keep the upper left corner and further transform the lower right corner. Multiplying the third row from the left by $\omega$ and then adding the second column, multiplied by $\alpha$ from the right, to the first column, we obtain in the lower right corner

$$\begin{aligned} \left( \begin{array}{ccc} \varvec{w} &{}\quad 0 &{}\quad \varvec{x} \\ 0 &{}\quad 1 &{}\quad 0 \\ - \varvec{\chi } - \bar{\omega } \alpha &{}\quad - \bar{\omega } &{}\quad - \omega \beta - \varvec{\psi } \\ \varvec{y} &{}\quad 0 &{}\quad \varvec{z} \end{array} \right) . \end{aligned}$$

Eliminating the entry $- \bar{\omega }$ and permuting, we obtain

We further transform the lower right $3 \times 2$ matrix. Multiplying its first row by $\alpha$ from the left, subtracting the second row and adding the third row, multiplied by $\beta$ from the left, the lower right $3 \times 2$ matrix becomes

$$\begin{aligned} \left( \begin{array}{cc} \alpha &{} \quad \beta \\ - \varvec{\chi } - \bar{\omega } \alpha &{} \quad - \omega \beta - \varvec{\psi } \\ \varvec{y} &{}\quad \varvec{z} \end{array} \right) , \end{aligned}$$

where we used the equations $\alpha \varvec{x} + (\omega \beta + \varvec{\psi }) + \beta \varvec{z} = \beta$ and $\alpha \varvec{w} + (\varvec{\chi } + \bar{\omega } \alpha ) + \beta \varvec{y} = \alpha$. Subtracting from the second column the first column, multiplied from the right by $\alpha ^{-1} \beta$, we obtain

$$\begin{aligned} \left( \begin{array}{cc} \alpha &{}\quad 0 \\ - \varvec{\chi } - \bar{\omega } \alpha &{} \quad - \omega \beta - \varvec{\psi } - (- \varvec{\chi } - \bar{\omega } \alpha ) \alpha ^{-1} \beta \\ \varvec{y} &{}\quad \varvec{z} - \varvec{y} \alpha ^{-1} \beta \end{array} \right) . \end{aligned}$$

Now eliminate the entries in the first column below $\alpha$, multiply the first column by $\alpha ^{-1}$, conjugate the second column by $\beta$ and multiply the second row by $-1$ to obtain

$$\begin{aligned} \left( \begin{array}{cc} 1 &{}\quad 0 \\ 0 &{}\quad \beta \omega + \beta \varvec{\psi } \beta ^{-1} + \beta (- \varvec{\chi } - \bar{\omega } \alpha ) \alpha ^{-1} \\ 0 &{} \quad \beta \varvec{z} \beta ^{-1} - \beta \varvec{y} \alpha ^{-1} \end{array} \right) . \end{aligned}$$

Adding the third row, multiplied by $1 + \beta$ from the left, to the second row yields

with $\tilde{j}_i$ as in (28) or (29).

All in all, we have transformed $\tilde{j'_i}$ by performing elementary column and row operations (which correspond to pre- and post-composition with isomorphisms) to $\tilde{j''_i}: \, \bigoplus _{\mathcal E^0 {\setminus } \left\{ v \right\} } K_i(A) \cong \mathrm{ker}\,(\phi _i) \rightarrow \bigoplus _{\mathcal E^0} K_i(A)$ given by

with $\tilde{j}_i$ as in (28) or (29). This completes the proof in the dihedral Artin–Tits case.

Let us now treat the torus knot case. With respect to the ordering of $\mathcal E^0$ as in (26), the original homomorphism $j_i$ in (20) is given by

(33)

where the row

is at the position corresponding to the finite word $a^{p-1}$ in (26), and the column

is at the position corresponding to the finite word ba in (26). By performing elementary row operations, which correspond to post-composition with isomorphisms, $j_i$ is transformed to

(34)

where

$$\begin{aligned} \varvec{w}= & {} - \beta - \alpha \beta - \alpha ^2 \beta - \cdots - \alpha ^{p-2} \beta , \ \varvec{x} = 1 - (\alpha + \beta \alpha + \beta ^2 \alpha \\&+ \cdots + \beta ^{q-2} \alpha )(\beta + \alpha \beta + \alpha ^2 \beta + \cdots + \alpha ^{p-2} \beta ). \end{aligned}$$

The map $\phi _i$ in (20) is given by $\phi = (\varvec{y}, 0, \dotsc , 0, \omega , \varvec{z}, 0, \dotsc , 0)$, where

$$\begin{aligned} \varvec{y}= & {} \beta \alpha + \beta ^2 \alpha + \cdots + \beta ^{q-1} \alpha , \ \omega = \alpha ^{p-1}, \ \varvec{z} = \alpha \beta + \alpha ^2 \beta + \cdots + \alpha ^{p-2} \beta , \end{aligned}$$

where $\omega$ is at the position corresponding to the finite word $a^{p-1}$ in (26) and $\varvec{z}$ is at the position corresponding to the finite word ba in (26).

Now the following map $\bigoplus _{\mathcal E^0 {\setminus } \left\{ a^{p-1} \right\} } K_i(A) \rightarrow \bigoplus _{\mathcal E^0} K_i(A)$ induces an isomorphism $\bigoplus _{\mathcal E^0 {\setminus } \left\{ a^{p-1} \right\} } K_i(A) \cong \mathrm{ker}\,(\phi _i)$:

where the double vertical line indicates the position of $a^{p-1}$. Pre-composing $j'_i$ by this isomorphism yields the homomorphism $\tilde{j'_i}: \, \bigoplus _{\mathcal E^0 {\setminus } \left\{ a^{p-1} \right\} } K_i(A) \cong \mathrm{ker}\,(\phi _i) \rightarrow \bigoplus _{\mathcal E^0} K_i(A)$ given by

Finally, by performing elementary column and row operations corresponding to pre- and post-composition with isomorphisms, $\tilde{j'_i}$ is transformed to $\tilde{j''_i}: \, \bigoplus _{\mathcal E^0 {\setminus } \left\{ a^{p-1} \right\} } K_i(A) \cong \mathrm{ker}\,(\phi _i) \rightarrow \bigoplus _{\mathcal E^0} K_i(A)$ given by

with $\tilde{j}_i$ as in (30). This completes the proof in the torus knot case. $\square$

Remark 5.7

The proof shows that $\tilde{j}_i$, $\tilde{\pi }_i$ and $\tilde{\partial }_i$ are related to the maps $j_i$, $\pi _i$ and $\partial _i \circ \psi _i^{-1} \circ q_i$ from (21) in the following way: We have isomorphisms $\mathrm{ker}\,(\phi _i) \cong K_i \oplus K_i(A)$ and $\bigoplus _{\mathcal E^0} K_i(A) \cong K_i \oplus (K_i(A) \oplus K_i(A))$ for some abelian group $K_i$, which fit into a commutative diagram

where the lower horizontal arrow is given by $\left( {\begin{matrix} * &{} 0 \\ 0 &{} \tilde{j}_i \end{matrix}} \right)$ for some automorphism $*: \, K_i \cong K_i$. Then $\tilde{\pi }_i$ is given by the composition

$$\begin{aligned} K_i(A) \oplus K_i(A) \hookrightarrow K_i \oplus (K_i(A) \oplus K_i(A)) \cong \bigoplus _{\mathcal E^0} K_i(A) \overset{\pi _i}{\longrightarrow } K_i(I), \end{aligned}$$

where the first map is the canonical inclusion, and $\tilde{\partial }_i$ is given by the composition

$$\begin{aligned} K_i(I) \overset{\partial _i \circ \psi _i^{-1} \circ q_i}{\longrightarrow } \mathrm{ker}\,(\phi _i) \cong K_i \oplus K_i(A) \twoheadrightarrow K_i(A), \end{aligned}$$

where the last map is the canonical projection. In particular, we have $\mathrm{im\,}(\tilde{\pi }_i) = \mathrm{im\,}(\pi _i)$.

5.4 Examples

We will use the following fact several times.

Remark 5.8

The commutator (or derived) subgroup $G'$ of a group G is the normal subgroup generated by all elements of the form $ghg^{-1}h^{-1}$ for $g,h\in G$. The quotient $G_{ab }=G/G'$ is called the abelianization of G and we let $f_{ab }:G\rightarrow G_{ab }$ denote the canonical quotient map. The following result can now be deduced from [39] (see Theorem 1.2 and the following paragraph in that paper): Suppose that G is a torsion-free group satisfying the Baum–Connes conjecture. Then there is a well-defined split-injective group homomorphism $G_{ab } \rightarrow K_1(C^*_r(G))$ defined by $f_{ab }(g)\mapsto [\lambda (g)]_1$.

Example 5.9

Let $A=\mathbb {C}$, so of course, the action is trivial, $K_0(\mathbb {C})=\mathbb {Z}$, and $K_1(\mathbb {C})=\{0\}$, and (27) gives the exact sequence

$$\begin{aligned} 0 \longrightarrow K_1(I) \overset{\tilde{\partial }_1}{\longrightarrow } \mathbb {Z}\overset{\tilde{j}_0}{\longrightarrow } \mathbb {Z}\oplus \mathbb {Z}\overset{\tilde{\pi }_0}{\longrightarrow } K_0(I) \longrightarrow 0. \end{aligned}$$

Consider the case $I_2(m)$ with $m\ge 2$ even. Then (28) gives that $\tilde{j}_0$ is the zero map, and consequently $K_1(I)\cong \mathbb {Z}$ and $K_0(I)\cong \mathbb {Z}^2$. Moreover, (22) now gives the exact sequence

$$\begin{aligned} 0 \longrightarrow K_1(C^*_r(I_2(m))) \longrightarrow K_0(I) \overset{\iota _0}{\longrightarrow } K_0(A) \longrightarrow K_0(C^*_{\lambda }(I_2(m))) \longrightarrow K_1(I) \longrightarrow 0. \end{aligned}$$

Since $\iota _0\circ \pi _0$ is the zero map (using the formula (23)), and $\pi _0$ is surjective, we get that $\iota _0$ is the zero map. Hence, $K_0(C^*_{\lambda }(I_2(m)))\cong K_0(A) \oplus K_1(I) \cong \mathbb {Z}^2$ and $K_1(C^*_{\lambda }(I_2(m)))\cong K_0(I) \cong \mathbb {Z}^2$. It follows from Remark 5.8 that the $K_1$-group is generated by $[\lambda (a)]$ and $[\lambda (b)]$. Furthermore, one copy of $\mathbb {Z}$ in $K_0$ is generated by [1].

Moreover, having computed the maps $j_i$ and $j'_i$ in (31) and (32), it is easy to determine the K-theory of $Q \mathrel {:=}C^*_r(G \ltimes \tilde{\Omega }_{\infty })$: We get $K_0(Q) \cong \mathbb {Z}\oplus (\mathbb {Z}/ \tfrac{m-2}{2} \mathbb {Z})$ and $K_1(Q) \cong \mathbb {Z}$. In case $m \ge 4$, Remark 4.6 tells us that this determines the stable UCT Kirchberg algebra Q up to isomorphism. Furthermore, we have exact sequences $0 \rightarrow J \rightarrow C^*_{\lambda }(P) \rightarrow C^*_{\lambda }(I_2(m)) \rightarrow 0$ and $0 \rightarrow \mathcal K(\ell ^2 P) \rightarrow J \rightarrow Q \rightarrow 0$.

Assume next that $m\ge 3$ is odd. Then (29) gives that $\tilde{j}_0$ is the map $\left( {\begin{matrix} 2 \\ 1 \end{matrix}}\right)$, which is injective with cokernel isomorphic to $\mathbb {Z}$. Therefore, $K_0(I)\cong \mathbb {Z}$ and $\tilde{\partial }_1$ is the zero map, so $K_1(I)=\{0\}$. From (22) we get the exact sequence

$$\begin{aligned} 0 \longrightarrow K_1(C^*_{\lambda }(I_2(m))) \longrightarrow K_0(I) \overset{\iota _0}{\longrightarrow } K_0(A) \longrightarrow K_0(C^*_{\lambda }(I_2(m))) \longrightarrow 0. \end{aligned}$$

Since $\iota _0\circ \pi _0$ is the zero map, and $\pi _0$ is surjective, then $\iota _0$ is the zero map. Hence, $K_0(C^*_{\lambda }(I_2(m)))\cong K_0(A)\cong \mathbb {Z}$ and thus $K_1(C^*_{\lambda }(I_2(m)))\cong K_0(I) \cong \mathbb {Z}$. The $K_0$-group is generated by the identity, and by Remark 5.8, the $K_1$-group is generated by $[\lambda (a)]=[\lambda (b)]$.

As above, using (31) and (32), it is easy to determine the K-theory of $Q \mathrel {:=}C^*_r(G \ltimes \tilde{\Omega }_{\infty })$: We get $K_0(Q) \cong \mathbb {Z}/ (m-2) \mathbb {Z}$ and $K_1(Q) \cong \left\{ 0 \right\}$. Remark 4.6 tells us that this determines the stable UCT Kirchberg algebra Q up to isomorphism, and we get $Q \cong \mathcal K\otimes \mathcal O_{m-1}$ (it is possible to write down an explicit isomorphism). So we have exact sequences $0 \rightarrow J \rightarrow C^*_{\lambda }(P) \rightarrow C^*_{\lambda }(I_2(m)) \rightarrow 0$ and $0 \rightarrow \mathcal K(\ell ^2 P) \rightarrow J \rightarrow \mathcal K\otimes \mathcal O_{m-1} \rightarrow 0$.

Finally, consider the torus knot group T(p, q) of type (p, q) for $p,q\ge 2$, and set $g=\gcd (p,q)$. Then (30) gives that $\tilde{j}_0$ is the map $\left( {\begin{matrix} p \\ q \end{matrix}}\right)$, which is injective with cokernel isomorphic to $\mathbb {Z}\oplus (\mathbb {Z}/ g \mathbb {Z})$. Therefore, $K_0(I)\cong \mathbb {Z}\oplus (\mathbb {Z}/ g\mathbb {Z})$ and $\tilde{\partial }_1$ is the zero map, so $K_1(I)=\{0\}$. From (22) we get the exact sequence

$$\begin{aligned} 0 \longrightarrow K_1(C^*_{\lambda }(T(p,q))) \longrightarrow K_0(I) \overset{\iota _0}{\longrightarrow } K_0(A) \longrightarrow K_0(C^*_{\lambda }(T(p,q))) \longrightarrow 0. \end{aligned}$$

Since $\iota _0\circ \pi _0$ is the zero map, and $\pi _0$ is surjective, then $\iota _0$ is the zero map. Hence, $K_0(C^*_{\lambda }(T(p,q)))\cong K_0(A)\cong \mathbb {Z}$ and $K_1(C^*_{\lambda }(T(p,q)))\cong K_0(I)\cong \mathbb {Z}\oplus (\mathbb {Z}/ g\mathbb {Z})$. The $K_0$-group is generated by the identity. To write down generators for $K_1$, let x and y be integers such that $\tfrac{p}{g} \cdot x - \tfrac{q}{g} \cdot y = 1$. Then $[\lambda (a^y b^{-x})]$ generates the direct summand $\mathbb {Z}$ and $[\lambda (a^{p/g} b^{-q/g})]$ generates the direct summand $\mathbb {Z}/ g \mathbb {Z}$ of $K_1$ (see Remark 5.8).

Having computed the maps $j_i$ and $j'_i$ in (33) and (34), it is easy to determine the K-theory of $Q \mathrel {:=}C^*_r(G \ltimes \tilde{\Omega }_{\infty })$: We get $K_0(Q) \cong \mathbb {Z}$, $K_1(Q) \cong \mathbb {Z}$ if $p = q = 2$, and $K_0(Q) \cong \mathbb {Z}/ ((p-1)(q-1) - 1) \mathbb {Z}$, $K_1(Q) \cong \left\{ 0 \right\}$ otherwise. In case $(p,q) \ne (2,2)$, Remark 4.6 tells us that this determines the stable UCT Kirchberg algebra Q up to isomorphism, and we get $Q \cong \mathcal K\otimes \mathcal O_{(p-1)(q-1)}$. Thus we obtain exact sequences $0 \rightarrow J \rightarrow C^*_{\lambda }(P) \rightarrow C^*_{\lambda }(T(p,q)) \rightarrow 0$ and $0 \rightarrow \mathcal K(\ell ^2 P) \rightarrow J \rightarrow \mathcal K\otimes \mathcal O_{(p-1)(q-1)} \rightarrow 0$.

Example 5.10

The braid group $B_4$ is given by the presentation

$$\begin{aligned} \langle a,b,c \, \vert \, aba=bab, bcb=cbc, ac=ca\rangle . \end{aligned}$$

Consider the surjective homomorphism $\varphi :B_4\rightarrow B_3$ given by $a\mapsto a$, $b\mapsto b$, $c\mapsto a$. Then it follows from [27, Theorem 2.1] that $\mathrm{ker}\,\varphi$ is isomorphic to the free group $\mathbb {F}_2$ with generators

$$\begin{aligned} x_1=ca^{-1} \quad \text {and}\quad x_2=bca^{-1}b^{-1}. \end{aligned}$$

Moreover, the obvious inclusion $B_3\hookrightarrow B_4$ defines an action $\gamma$ of $B_3$ on $\mathbb {F}_2$ given by

$$\begin{aligned} {\begin{matrix} \gamma _a(x_1)=ax_1a^{-1}=x_1, &{}\quad \gamma _b(x_1)=bx_1b^{-1}=x_2, \\ \gamma _a(x_2)=ax_2a^{-1}=x_1^{-1}x_2, &{}\quad \gamma _b(x_2)=bx_2b^{-1}=x_2x_1^{-1}x_2, \end{matrix}} \end{aligned}$$

and $B_4\cong \mathbb {F}_2\rtimes _\gamma B_3$.

Set $A=C^*_{\lambda }(\mathbb {F}_2)$, and note that $C^*_{\lambda }(B_4)\cong A\rtimes _r B_3$, where the action is induced from the one above. It is shown in [44, Corollary 3.2] that the K-theory of $C^*_{\lambda }(\mathbb {F}_2)$ is given by $K_0(C^*_{\lambda }(\mathbb {F}_2))=\mathbb {Z}[1]$ and $K_1(C^*_{\lambda }(\mathbb {F}_2))=\mathbb {Z}[\lambda (x_1)] \oplus \mathbb {Z}[\lambda (x_2)]$, where $\lambda$ is the left regular representation Hence, it follows that $\alpha _0=\beta _0=id$ on $K_0$, while on $K_1$ we have

$$\begin{aligned} \alpha _1=\begin{pmatrix} 1 &{} -1 \\ 0 &{} 1 \end{pmatrix}^{-1}=\begin{pmatrix} 1 &{} 1 \\ 0 &{} 1 \end{pmatrix} \quad \text {and}\quad \beta _1=\begin{pmatrix} 0 &{} -1 \\ 1 &{} 2 \end{pmatrix}^{-1}=\begin{pmatrix} 2 &{} 1 \\ -1 &{} 0 \end{pmatrix}. \end{aligned}$$

Thus, using (29), we compute that

$$\begin{aligned} \tilde{j}_0=\begin{pmatrix} 2 \\ 1 \end{pmatrix} \quad \text {and}\quad \tilde{j}_1=\begin{pmatrix} 2 &{} 2 \\ -1 &{} 0 \\ 1 &{} 2 \\ 0 &{} 0 \end{pmatrix}. \end{aligned}$$

Both maps are injective, with cokernels isomorphic to $\mathbb {Z}$ and $\mathbb {Z}^2\oplus (\mathbb {Z}/ 2\mathbb {Z})$, respectively. This means that the index maps are trivial, so (27) gives short exact sequences

$$\begin{aligned} 0 \longrightarrow K_i(A) \overset{\tilde{j}_i}{\longrightarrow } K_i(A) \oplus K_i(A) \overset{\tilde{\pi }_i}{\longrightarrow } K_i(I) \longrightarrow 0 \end{aligned}$$

for $i=0,1$. Thus, $K_0(I)=\mathbb {Z}$ and $K_1(I)=\mathbb {Z}^2\oplus (\mathbb {Z}/ 2\mathbb {Z})$.

Moreover, using the formula (23), we see that $\iota _i\circ \pi _i:K_i(A) \oplus K_i(A) \oplus K_i(A) \oplus K_i(A) \rightarrow K_i(A)$ is the zero map for $i=0$ and

for $i=1$ (with respect to $\mathcal E^0 = \left\{ aa,ab,ba,bb \right\}$ with this ordering). In both cases, $\pi _i$ is surjective (as $\mathrm{im\,}(\pi _i) = \mathrm{im\,}(\tilde{\pi }_i)$ by Remark 5.7), so $\iota _0=0$, and $\iota _1$ maps $\mathbb {Z}^2\oplus (\mathbb {Z}/ 2\mathbb {Z})$ onto $\mathbb {Z}^2$. Therefore, (22) gives an exact sequence

$$\begin{aligned} 0 \longrightarrow K_0(A) \longrightarrow K_0(A\rtimes _r B_3) \longrightarrow K_1(I) \overset{\iota _1}{\longrightarrow } K_1(A) \longrightarrow K_1(A\rtimes _r B_3) \longrightarrow K_0(I) \longrightarrow 0, \end{aligned}$$

that is,

$$\begin{aligned} 0 \longrightarrow \mathbb {Z}\longrightarrow K_0(A\rtimes _r B_3) \longrightarrow \mathbb {Z}^2\oplus (\mathbb {Z}/ 2\mathbb {Z}) \overset{\iota _1}{\longrightarrow } \mathbb {Z}^2 \longrightarrow K_1(A\rtimes _r B_3) \longrightarrow \mathbb {Z}\longrightarrow 0. \end{aligned}$$

Since $\iota _1$ is surjective, $K_1(C^*_{\lambda }(B_4))\cong K_0(I) \cong \mathbb {Z}$, while $K_0(C^*_{\lambda }(B_4))$ is either $\mathbb {Z}$ or $\mathbb {Z}\oplus (\mathbb {Z}/ 2\mathbb {Z})$. It is known that [1] must generate a copy of $\mathbb {Z}$ as a direct summand of the $K_0$-group (see for instance [31, Proof of Lemma 4.4]), so therefore it follows that $K_0(C^*_{\lambda }(B_4))\cong \mathbb {Z}\oplus (\mathbb {Z}/ 2\mathbb {Z})$. Moreover, it follows from Remark 5.8 that $K_1$ is generated by $[\lambda (a)]=[\lambda (b)]=[\lambda (c)]$.

Example 5.11

Artin’s representation of braid groups (see [42] for a discussion) is defined for $n\ge 3$ as the action $\gamma$ of $B_n$ on $\mathbb {F}_n$, with canonical generators $\sigma _i$ and $x_i$, respectively, given by

$$\begin{aligned} \gamma (\sigma _i)(x_j)= {\left\{ \begin{array}{ll} x_{i+1} &{}\quad \text {if } j=i, \\ x_{i+1}^{-1}x_i x_{i+1} &{}\quad \text {if } j=i+1, \\ x_j &{}\quad \text {else}. \end{array}\right. } \end{aligned}$$

We consider the case $n=3$, and want to compute K-theory for the reduced group C*-algebra of $G = \mathbb {F}_3 \rtimes B_3$. Note that by Crisp and Paris [12, Proposition 2.1], G is the irreducible Artin–Tits group given by the presentation

$$\begin{aligned} \left\langle a, b, c \, \vert \, aba=bab, bcbc=cbcb, ac=ca \right\rangle . \end{aligned}$$

Let us compute the induced action (denoted by the same symbol) $\gamma$ of $B_3$, now generated by a, b, on $A = C^*_{\lambda }(\mathbb {F}_3)$ in K-theory. We get $\alpha _0=\beta _0=id$ on $K_0$, while on $K_1$ we have

$$\begin{aligned} \alpha _1=\begin{pmatrix} 0 &{} 1 &{} 0 \\ 1 &{} 0 &{} 0 \\ 0 &{} 0 &{} 1 \end{pmatrix}, \quad \beta _1=\begin{pmatrix} 1 &{} 0 &{} 0 \\ 0 &{} 0 &{} 1 \\ 0 &{} 1 &{} 0 \end{pmatrix}, \quad \text {and}\quad (\gamma _{aba=bab})^{-1}_*=\begin{pmatrix} 0 &{} 0 &{} 1 \\ 0 &{} 1 &{} 0 \\ 1 &{} 0 &{} 0 \end{pmatrix}. \end{aligned}$$

Thus, using (29), we compute that

$$\begin{aligned} \tilde{j}_0=\begin{pmatrix} 2 \\ 1 \end{pmatrix} \quad \text {and}\quad \tilde{j}_1=\begin{pmatrix} 1 &{} 0 &{} 1 \\ 0 &{} 2 &{} 0 \\ 1 &{} 0 &{} 1 \\ 0 &{} 1 &{} 0 \\ 0 &{} 1 &{} 0 \\ 1 &{} -1 &{} 1 \end{pmatrix}. \end{aligned}$$

We note that $\mathrm{ker}\,\tilde{j}_1\cong \mathbb {Z}$, ${\text {im}}\tilde{j}_1\cong \mathbb {Z}^2$, and ${\text {coker}}\tilde{j}_1\cong \mathbb {Z}^4$. Since $\tilde{j}_0$ is injective, we get the exact sequence

$$\begin{aligned}&0 \longrightarrow K_0(A) \overset{\tilde{j}_0}{\longrightarrow } K_0(A)\oplus K_0(A) \overset{\tilde{\pi }_0}{\longrightarrow } K_0(I) \overset{\tilde{\partial }_0}{\longrightarrow }\\&\quad K_1(A) \overset{\tilde{j}_1}{\longrightarrow } K_1(A)\oplus K_1(A) \overset{\tilde{\pi }_1}{\longrightarrow } K_1(I) \longrightarrow 0, \end{aligned}$$

which gives

$$\begin{aligned} 0 \longrightarrow \mathbb {Z}\overset{\tilde{j}_0}{\longrightarrow } \mathbb {Z}^2 \overset{\tilde{\pi }_0}{\longrightarrow } K_0(I) \overset{\tilde{\partial }_0}{\longrightarrow } \mathbb {Z}^3 \overset{\tilde{j}_1}{\longrightarrow } \mathbb {Z}^6 \overset{\tilde{\pi }_1}{\longrightarrow } K_1(I) \longrightarrow 0. \end{aligned}$$

Here, $K_0(I)\cong {\text {coker}}\tilde{j}_0\oplus \mathrm{ker}\,\tilde{j}_1\cong \mathbb {Z}^2$ and $K_1(I)\cong {\text {coker}}\tilde{j}_1\cong \mathbb {Z}^4$. Moreover, using the formula (23), we see that $\iota _i\circ \pi _i:K_i(A) \oplus K_i(A) \oplus K_i(A) \oplus K_i(A) \rightarrow K_i(A)$ is the zero map for $i=0$ and

for $i=1$ (with respect to $\mathcal E^0 = \left\{ aa,ab,ba,bb \right\}$ with this ordering). Since $\pi _1$ is surjective ($\mathrm{im\,}(\pi _1) = \mathrm{im\,}(\tilde{\pi }_1)$ by Remark 5.7), we compute that for $\iota _1:\mathbb {Z}^4\rightarrow \mathbb {Z}^3$ that $\mathrm{ker}\,\iota _1\cong \mathbb {Z}^2$, ${\text {im}}\iota _1\cong \mathbb {Z}^2$, and ${\text {coker}}\iota _1\cong \mathbb {Z}$. On the other hand, the image of $\pi _0$ is $\mathbb {Z}$ (as $\mathrm{im\,}(\pi _0) = \mathrm{im\,}(\tilde{\pi }_0)$ by Remark 5.7), so the kernel of $\iota _0$ is either $\mathbb {Z}$ or $\mathbb {Z}^2$. If the former holds, then $\iota _0$ is not zero, so that $[1] \in K_0(A \rtimes _r B_3)$ must be a torsion element, which is not possible, because the canonical trace does not vanish on [1]. Thus, we must have that the kernel is $\mathbb {Z}^2$, which means that $\iota _0$ is the zero map, so we get

$$\begin{aligned} 0\longrightarrow & {} K_0(A) \longrightarrow K_0(A\rtimes _r B_3) \longrightarrow K_1(I) \overset{\iota _1}{\longrightarrow } K_1(A) \longrightarrow K_1(A\rtimes _r B_3) \\\longrightarrow & {} K_0(I) \longrightarrow 0, \end{aligned}$$

which gives

$$\begin{aligned} 0\longrightarrow & {} \mathbb {Z}\longrightarrow K_0(A\rtimes _r B_3) \longrightarrow \mathbb {Z}^4 \overset{\iota _1}{\longrightarrow } \mathbb {Z}^3\\\longrightarrow & {} K_1(A\rtimes _r B_3) \longrightarrow \mathbb {Z}^2 \longrightarrow 0. \end{aligned}$$

Hence, $K_0(C^*_{\lambda }(G)) \cong K_0(A\rtimes _r B_3)\cong \mathbb {Z}^3$ and $K_1(C^*_{\lambda }(G)) \cong K_1(A\rtimes _r B_3)\cong \mathbb {Z}^3$. The identity generates one copy of $\mathbb {Z}$ in $K_0$, while $[\lambda (a)]=[\lambda (b)]$ and $[\lambda (c)]$ generate two copies of $\mathbb {Z}$ in $K_1$ by Remark 5.8.

Remark 5.12

In all our examples, we have $K_0(C^*_r(G))\cong H_0G\oplus H_2G$ and $K_1(C^*_r(G))\cong H_1G\oplus H_3G$. The reason is the following: As explained in the introduction of [19], there is a manifold model for the classifying space BG with dimension at most 3 in all our examples, because we are considering Artin–Tits groups of finite type with rank at most 3. Thus, by Matthey [38, Proposition 2.1. (ii)], we have $K_*(C^*_{\lambda }(G)) \cong K_*^G(\underline{E}G) = K_*^G(EG) \cong K_*(BG) \cong H_*(G) \oplus H_{*+2}(G)$. For the first identification, we used that G satisfies the Baum–Connes conjecture, while for the second equality, we used that G is torsion-free.