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

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.

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 [8,9,22] 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 5.5).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 § 5.4).
The main ingredient for our computations is a K-theory formula for crossed products in the presence of the independence condition (see [13,14]).Therefore, our K-theory results point towards the possibility of using Ktheory formulas for semigroup C*-algebras and certain crossed products as in [13,14] 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 [11]).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.

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2.1.Monoids defined by presentations.Let Σ be a countable set, Σ * the free monoid generated by Σ (also viewed as the set of finite words in Σ) and R a subset of Σ * × Σ * .We call (Σ, R) a presentation, where Σ is the set of generators and R is the set of relations.Words u ∈ Σ * for which (u, v) or (v, u) lies in R are called relators.Given a presentation (Σ, R), we form the monoid P = Σ | R + generated by Σ subject to the relations u = v for all (u, v) ∈ R (see [7, § 1.12]).Given x ∈ Σ * , we also denote by x the element in P represented by x.Given x, y ∈ Σ * , we write x ≡ y if x and y coincide in Σ * , i.e., x is the same word as y, and x = y if x and y represent the same element in P. We write ℓ * for the length of a word in Σ * and ℓ(x) := min {ℓ * (w): w ∈ Σ * , w = x} for the word length of x ∈ P with respect to Σ.Given σ ∈ Σ and x ∈ Σ * , ℓ σ (x) counts how many times σ appears in x.We write ε for the empty word in Σ * .
Recall how P = Σ | R + is constructed: Two words x, y ∈ Σ * represent the same element in P if and only if there is a finite sequence w 1 , . . ., w s ∈ Σ * with w 1 ≡ x, w s ≡ y, and for all 1 ≤ i ≤ s − 1, there exist a, u, v, z ∈ Σ * with (u, v) ∈ R or (v, u) ∈ R or (u, v) = (ε, ε) such that w i ≡ auz and w i+1 ≡ avz.
As an immediate consequence, it follows that if for all (u, v) ∈ R, we have u ε v, then 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 * = {e}, where e is the identity in P.
In the following, we will exclude the (degenerate) case that a generator σ ∈ Σ is redundant, i.e., we have a relation of the form (σ, w) or (w, σ) where w is a word not involving σ (in other words, w ∈ (Σ \ {σ}) * ).In that case, we can give another presentation for P, P Σ \ {σ} | R ′ + , where R ′ is obtained from R by deleting (σ, w) or (w, σ) and replacing every (u, v) ∈ R by (u ′ , v ′ ), where we obtain u ′ from u and v ′ from v by replacing every σ in u and v by w.Moreover, we certainly can assume that R does not contain (u, u) for any u ∈ Σ * .
We will mainly focus on the case of one-relator monoids, i.e., monoids of the form P = Σ | R + where R consists of a single pair (u, v) ∈ Σ * × Σ * .This is (arguably) the most basic case.Moreover, in examples, we actually see very different phenomena when we drop this assumption (see § 4.2).We often write (Σ, u = v) for (Σ, R) and Σ | u = v + for Σ | R + .Let G = Σ | u = v be the group generated by Σ 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 → G induced by the identity on Σ is injective.In particular, P is cancellative (i.e., left and right cancellative).Conversely, since we assume that u v, if P is cancellative, then the first and last letters of u and v must differ.
Remark 2.1.G = Σ | u = v is a one-relator group.If the first and last letters of u and v differ, then it follows from [34,Theorem 4.12] that G is torsion-free.Moreover, G is amenable if and only if G is cyclic or G a, b | ab = b k a for some 0 k ∈ Z (see for instance [6]).
2.1.1.The LCM property.We will further assume that our one-relator monoid P is right LCM, i.e., given p, q ∈ P, we have pP ∩ qP = ∅ or there exists r ∈ P with pP ∩ qP = rP.
In order to deduce the strong r-cube condition (on Σ * ), we need to check whether (Σ, u = v) is r-homogeneous in the sense of [18,Definition 4.1].This is for instance the case if ℓ * (u) = ℓ * (v) (for we can take Then define λ as the monoid homomorphism from Σ * to the non-negative integers by setting λ(σ) = ζ and λ(τ) = η for all τ ∈ Σ \ {σ}.We then have λ [18,Definition 2.1].This in turn implies that P = Σ, u = v + is right LCM by [18,Proposition 6.10].we cannot find t, w ∈ Σ * with tuw ≡ tvw such that the first letter of tuw is σ and the first letter of tvw is τ (or vice versa).
Moreover, let us present examples of presentations ({a, b} , u = v) which do not give rise to left reversible monoids.Define OVL(v) = x ∈ {a, b} * : v ≡ xy ≡ wx for some ε w, y ∈ Σ * .Assume that OVL(v) = {ε}.Further suppose that ℓ a (u) < ℓ a (v) or ℓ b (u) < ℓ b (v).The first condition implies that ({a, b} , u = v) satisfies the Church-Rosser condition (by [46]), while the second condition ensures that ({a, b} , u = v) is noetherian.(The reader may consult [4, § 1.1] for explanations of these notions.)This implies that given w ∈ Σ * , we can find a unique irreducible word x ∈ Σ * (i.e., x does not contain v as a subword) such that w = x in P, and two irreducible words x, y ∈ Σ * represent the same element in P if and only if x ≡ y (see [4,Theorem 1.1.12]).Now assume that OVL(v) = {ε}, and that ℓ a (u) < ℓ a (v) or ℓ b (u) < ℓ b (v).If the first letter of v is a, but v ab k for some k ≥ 1, then P is not left reversible: Let p = b m , q = ab n for some m, n ≥ ℓ * (v).Take x, y ∈ Σ * irreducible.Then px and qy are irreducible, but px qy as px qy.Hence pP ∩ qP = ∅.
Further suppose that P embeds into a group G. Fix such an embedding and view P as a submonoid of G. Then there is a partial action G Ω P determined by the partial homeomorphisms U g −1 → U g , χ → g. χ, where U g −1 is non-empty only if g = pq −1 for some p, q ∈ P, and if g is of the form g = pq −1 , then [15,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 Ω P as well as isomorphic to the reduced C*-algebra of the corresponding partial transformation groupoid, For brevity, let us drop indices and write J := J P , J × = J \ {∅} and Ω := Ω P .Let us describe elements in Ω.
First of all, every pP ∈ J determines a point χ pP ∈ Ω given by χ pP (xP) = 1 if pP ⊆ xP and χ pP (xP) = 0 if pP xP.This allows us to identify J × with a subset of Ω.It is easy to see that J × is dense in Ω.We define Ω ∞ = Ω \ J × .Clearly, J × and Ω ∞ are G-invariant subspaces.In the case where P has no invertible elements other than the identity, we have a bijection P J × , p → pP, so that we will not distinguish between P and J × , and hence we have Ω ∞ = Ω \ P.Among the points in Ω ∞ , we single out those χ ∈ Ω for which χ −1 (1) is maximal, i.e., whenever ω ∈ Ω satisfies ω(xP) = 1 for all xP ∈ J with χ(xP) = 1, then we must have ω = χ.We set Ω max := χ ∈ Ω: χ −1 (1) is maximal .Note that Ω max ⊆ Ω ∞ .Moreover, we define ∂Ω := Ω max .Let us now collect a few facts about ∂Ω, which are obtained in [15, § 5.7] in greater generality than needed here.∂Ω is the minimal non-empty closed G-invariant subspace of Ω.Moreover, ∂Ω reduces to a single point (namely χ ∈ Ω given by χ(xP) = 1 for all x ∈ P, we usually denote this χ by ∞) if and only if Now let us assume that our right LCM monoid P is given by a presentation (Σ, R) as in § 2.1.Let G be the group given by the same presentation as P, and suppose that P → G induced by id Σ is an embedding (this must be the case if P embeds into a group).Moreover, we assume that for all (u, v) ∈ R, we have u ε v, so that P * = {e} and we may write P for the subset J of Ω.In this setting, let us describe Ω ∞ by infinite words (with letters) in Σ.
We write Σ ∞ for the set of all these infinite words.Given w ∈ Σ ∞ , define [w] i to be the word consisting of the first i letters of w (for i ≥ 1).In this picture with infinite words, the G-action on Ω ∞ is given as follows: As we explained above U g −1 is only non-empty if g = pq −1 for some p, q ∈ Σ * , and in that case, χ ∈ Ω ∞ lies in U g −1 if and only if χ = χ w for some w ∈ Σ ∞ which starts with q ∈ Σ * , i.e., of the form w = qx for some x ∈ Σ ∞ , and then g. χ = g.χ w = χ px .
The topology on Ω ∞ is the subspace topology from Ω, which in turn is determined by the basic open subspaces U(pP; p 1 P, . . ., , where p j P pP for all 1 ≤ j ≤ k.So basic open subspaces of Ω ∞ are given by Ω ∞ ∩ U(pP; p 1 P, . . ., p k P).
In general, two different infinite words can give rise to the same element in Ω ∞ .However, this cannot happen for infinite words which contain no relator as a finite subword: Lemma 2.4.For w, w ∈ Σ ∞ , let w i := [w] i and wj := [w] j for all i, j ≥ 1.Then χ w = χ w if and only if for all i, there exists j with w i ∈ wj P, and for all j, there exists i with wj ∈ w i P.
Assume that the infinite word w does not contain any relator as a finite subword.Then χ w = χ w if and only if w ≡ w.
Proof.The first part follows immediately from the construction of characters from infinite words.For the second part, χ w = χ w implies that w i = wj x for some x ∈ Σ * .Since w i contains no relator as a subword, we must have w i ≡ wj x, and hence w i ≡ wi .

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Throughout this section, let P be a right LCM monoid given by a presentation (Σ, R) as in § 2.1.Let G be the group given by the same presentation as P, and suppose that P → G induced by id Σ is an embedding.Moreover, we assume that every relator u satisfies u ε, so that P * = {e}.Also, recall that we do not allow generators in Σ to be redundant, and R does not contain (u, u) for any u ∈ Σ * .Note that in this situation, P is finitely generated if and only if |Σ| < ∞.
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 ∂Ω is as large as it can be.
, with p j P pP for all 1 ≤ j ≤ k.In case P is finitely generated, we may assume that Ω ∞ ∩ U ∅ and need to find g ∈ G with g. χ ∈ U (g. χ will then automatically lie in Ω ∞ ∩ U because χ ∈ Ω ∞ and Ω ∞ is G-invariant).In case P is not finitely generated, we only know that U ∅, and we need to find g ∈ G with g. χ ∈ U.
Let us first of all show that in the second case, we actually also have Ω ∞ ∩ U ∅. Choose generators σ j ∈ Σ such that p j P ⊆ pσ j P. As Σ is infinite, there exists σ ∈ Σ with σ j σ j P. Consider the infinite word w = pσzzz . . ., and the corresponding element χ w .Obviously, χ w (pP) = 1.If χ w (p j P) = 1, then pσz m ∈ p j P for some m, which implies that σz m ∈ σ j P.So there exists x ∈ Σ * with σz m = σ j x.But by assumption, σz m contains no relator as a subword.Hence we must have σz m ≡ σ j x.This is a contradiction as σ σ j .Now choose pj ∈ Σ * with pj = p j in P. We claim that regardless whether P is finitely generated or not, we can always find x ∈ pP with x p j P for all j and ℓ(x) ≥ ℓ * ( pj ) for all j.In the second case where P is not finitely generated, this follows from our argument above and because lim m→∞ ℓ(pσz m ) = ∞.In the first case where P is finitely generated, we can find t ∈ Σ ∞ with χ t ∈ U because Ω ∞ ∩U ∅.Set t i := [t] i .Then t 1 P t 2 P t 3 P . . ., which implies that sup i ℓ(t i ) = ∞.Otherwise {t i : i = 1, 2, . ..} would be a finite set since Σ is finite, but then we could not have t 1 P t 2 P t 3 P . . . .Now choose i with ℓ(t i ) ≥ max j ℓ * ( pj ) and set x := t i .
So let x ∈ pP satisfy x p j P for all j and ℓ(x) ≥ ℓ * ( pj ) for all j.Set g = xz.We have to show that g. χ ∈ U. Take s ∈ Σ ∞ with χ = χ s , and set s i := [s] i .Then g. χ = g.χ s = χ xzs .Since x ∈ pP, we have (g.χ)(pP) = 1.Assume that (g.χ)(p j P) = 1 for some j.Then there exists n with xzs n = pj r for some r ∈ Σ * .This means that we can find y 1 , . . ., y m ∈ Σ * with y 1 ≡ xzs n , y m ≡ pj r and for all 1 ) such that y l ≡ αuζ and y l+1 ≡ αvζ.By our assumptions on z, the subword z of xzs n will not be changed.Hence for every j, we must have y l ≡ x l zs n,l , with x l = x and s n,l = s n in P. In particular, we have x m zs n,m ≡ pj r.Now ℓ * (x m ) ≥ ℓ(x) ≥ ℓ * ( pj ), so that x m ≡ pj q for some q ∈ Σ * .This implies x ∈ pj P = p j P, which is a contradiction.This shows g. χ ∈ U. Now let us find ω ∈ Ω ∞ with trivial stabilizer group.Our assumption implies that |Σ| ≥ 2. Hence we can choose a sequence σ 1 , σ 2 , . . . in Σ such that the infinite word σ 1 σ 2 . . .has no period.Now form the infinite word w := zσ 1 zσ 2 z . . . .It has no period, either, because we cannot shift by a multiple of ℓ * (z) + 1 as σ 1 σ 2 . . .has no period, and if shifting w by a different number of letters produces w again, then all the σ i would have to coincide with a fixed letter of z, which again is not possible as σ 1 σ 2 . . .has no period.By our assumption on z, w does not contain a relator as a finite subword.Now assume that h ∈ G satisfies h.χ w = χ w .Then we must have h = pq −1 , and χ w (qP) = 1.The latter implies that zσ 1 zσ 2 . . .zσ n ∈ qP for some n, so that zσ 1 zσ 2 . . .zσ n = qr in P.But since zσ 1 zσ 2 . . .zσ n contains no relator as a subword, we must have zσ 1 zσ 2 . . .zσ n ≡ qr.In particular, w ≡ q w for some w ∈ Σ ∞ , and w also contains no relator as a finite subword and has no period.Now we have χ w = h.χ w = h.χ q w = χ p w .Since w contains no relator as a finite subword, Lemma 2.4 implies that q w ≡ w ≡ p w. Since w has no period, we must have ℓ * (p) = ℓ * (q), which implies p ≡ q.Hence h = pq −1 = e.
The following is an immediate consequence:

Corollary 3.2. Let us keep the same assumptions as in Theorem 3.1. Then if our semigroup P is finitely generated, we have ∂Ω
In both cases, i.e., regardless whether P is finitely generated or not (and G does not need to be exact), ∂C * λ (P) is a purely infinite simple C*-algebra.Here we used [23,Theorem 22.9].The last part of the corollary follows from Theorem 3.1 together with [15, Corollary 5.7.17].
Let us now apply our findings to the case of right LCM monoids P = Σ | u = v + determined by one defining relation.Recall that we do not allow a generator in Σ to be redundant, and that we assume that the first letter of u is not equal to the first letter of v and the last letter of u is not equal to the last letter of v.
Then there exists ε z ∈ Σ * 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 ∈ Σ be a generator which is not the first letter of u or v, and let a ∈ Σ be a generator which is not the last letter of u or v. Consider a i c j for i, j > max(ℓ * (u), ℓ * (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 ≡ a k c l with k, l ≥ 1.By our assumptions, this means that v does not start with a and does not end with c. Choose b ∈ Σ with b a, b c.Then a i bc j has all the desired properties.[15, § 5.8] if and only if P ⊆ G is quasi-lattice ordered in Nica's sense [38], i.e., for every g ∈ G, P ∩ gP = ∅ or there exists p ∈ P with P ∩ gP = pP.
and all the projections 1 P , 1 σ P and 1 uP are Murray-von Neumann equivalent to 1 in C * λ (P).Hence Let us discuss another natural topic, namely nuclearity of ∂C * λ (P), where Let us now present conditions on the presentation (Σ, u = v) which ensure that ∂C * λ (P) is nuclear.For this purpose, we introduce the following u is not a subword of v, no prefix of u is a suffix of v, and no suffix of u is a prefix of v.
We now construct graph models for G ∂Ω.
1} as the set of vertices, and introduce an edge µ with r(µ) = y and s(µ) = x if and only if there are σ, τ ∈ Σ with yτ ≡ σ x v.Here r and s are the range and source maps.
Ω ∞ is given as follows: Let Σ R be the finite subset of Σ containing all the letters that appear in the relators u and v.As the vertex set of our graph model, take Introduce an edge µ with r(µ) = y and s(µ) = x if there are σ, τ ∈ Σ R with yτ ≡ σ x v, and for every vertex v and every σ ∈ Σ \ Σ R , introduce a loop µ σ,v at v. Note that 3 ≤ |Σ| ensures that our monoid is not left reversible (see § 2.1.2).

Proof. (i):
We denote the graph by E, and write E 1 for the set of edges.By construction, we have a map σ : E 1 → Σ sending µ to the first letter of r(µ).σ extends to maps on the finite path space , where ∂E is the unit space of G E as in [31, § 3.2] (in our case, we have ∂E = E ∞ ).Let us show that the map As explained in § 2.1, our conditions ( 1) and ( 2) imply that for every z ∈ Σ * , we can find a unique irreducible word x ∈ Σ * (i.e., x does not contain v as a subword) such that z = x in P, and two irreducible words x, y ∈ Σ * represent the same element in P if and only if x ≡ y.Given z ∈ Σ * , we write ρ(z) for the unique irreducible word in Σ * with ρ(z) = z in P. Given χ ∈ Ω ∞ , Lemma 2.3 implies that we can find x ∈ Σ ∞ such that χ = χ x .Our assumption that OVL(v) = {ε} implies that x is of the form x = y 1 vy 2 vy 3 v . . ., where y i ∈ Σ * do not contain v as a subword.Set w := y 1 uy 2 uy 3 u . . . .Then w does not contain v as a finite subword by our assumptions on u and v, and we have χ = χ x = χ w by Lemma 2.4 (since x i = w i for infinitely many i).Moreover, given infinite words w, w ∈ Σ ∞ which do not contain v as finite subword such that χ w = χ w , then Lemma 2.4 implies that for all j ≥ l := ℓ * (v), we have w i ∈ wj P for some i, where w i = [w] i and wj = [ w] j .By our assumption on u and v, we conclude that i ≥ j − l and w j−l ≡ wj−l .Hence w = w.Since our graph E is constructed so that E ∞ consists of precisely those infinite words in Σ ∞ which do not contain v as a finite subword, it is clear that the map ϕ constructed above, ϕ : It is easy to see that it is a homeomorphism.Finally, it follows directly from our assumptions on u and v that (ii): It is straightforward to carry over the proof in case (i).Again, let our graph be E, and write E 1 for the set of edges.Define a map σ : E 1 → Σ as follows: For a vertex µ with r(µ) = y and s(µ) = x, where there are σ, τ ∈ Σ R with yτ ≡ σ x v, let σ(µ) be the first letter of r(µ) = y.For a vertex v and σ ∈ Σ \ Σ R , set σ(µ σ,v ) = σ.Extend σ to E * and E ∞ .Then it is straightforward to check that ϕ : ∂E → Ω, µ → χ σ(µ) is a homeomorphism and extends to an isomorphism of topological groupoids

T
Throughout this section, let P be left reversible.In that case, ∂Ω reduces to a point, given by the map J → {0, 1} sending every non-empty set in J to 1.We denote this point by ∞.As above, let us assume that P is right LCM.Recall that Ω ∞ = Ω \ P. In the following, we write Ω := Ω \ {∞} and Ω∞ := Ω ∞ \ {∞}.

4.1.
Reversible one-relator monoids.Let us present one-relator monoids P = Σ | u = v + which are left reversible and for which we can describe G ⋉ Ω∞ .Recall that we do not allow a generator in Σ to be redundant, and that we assume that the first letter of u is not equal to the first letter of v and the last letter of u is not equal to the last letter of v.
As explained in § 2.1, P = Σ | u = v + can only be left reversible if |Σ| ≤ 2. Since our assumptions force |Σ| ≥ 2, we must have |Σ| = 2, say Σ = {a, b}.The next lemma gives a special condition which enforces left reversibility.(ii) Again, take x ∈ Σ * and proceed inductively on ℓ * (x).Our claim certainly holds for x = ε.Assume w i ∈ xP, say w i = xy for some y ∈ P. By (i), we know that yw = wz for some z ∈ P.
For instance, if P is a Garside monoid in the sense of [20,17,19], 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 ∈ P with w = aα = αγ = bβ = βδ for some α, β, γ, δ ∈ P, then χ ∈ Ω satisfies χ(pP) = 1 for all p ∈ P if and only if χ( Then Ω admits the following decomposition as a (set-theoretic) disjoint union We write X := X 0 .Our goal is to describe As the vertex set, take the (finite) collection of all finite subwords in relator-free infinite words of length at most l.Introduce an edge µ with r(µ) = y ∈ E 0 and s(µ) = x ∈ E 0 if and only if there are σ, τ ∈ {a, b} with yτ ≡ σ x, and no relator is a subword of yτ ≡ σ x. 2. Assume that w in Lemma 4.1 satisfies w u and w v. Suppose that 2.1.sup ℓ * (x): x ∈ {a, b} * , x = w in P < ∞ (i.e., there are only finitely many words in {a, b} * representing w); 2.2.u is not a subword of v, that no prefix of u is a suffix of v, and that no suffix of u is a prefix of v; 2.3.there exists L ≥ 0 such that for all z = ar ∈ {a, b} * and z = bs ∈ {a, b} * with no v as a subword and with ℓ * (r), ℓ * (s) ≥ L, we must have zP ∩ zP ⊆ wP.Let W := x ∈ {a, b} * : x = w in P and l := max({ℓ * (z): z ∈ {v} ∪ W }) − 1.Then a graph model for (G ⋉ Ω∞ ) | Y is given as follows: As the vertex set, take the (finite) collection of all finite subwords in v, w-free infinite words of length at most l.Introduce an edge µ with r(µ) = y ∈ E 0 and s(µ) = x ∈ E 0 if and only if there are σ, τ ∈ {a, b} with yτ ≡ σ x, and no z ∈ {v} ∪ W is a subword of yτ ≡ σ x.
Here an infinite word is called relator-free if it does not contain a relator as a finite subword, and v-free if it does not contain v as a finite subword.An infinite word is called v, w-free if it is v-free and contains no w ∈ {a, b} * as a subword with w = w in P.
In case 2., note that as in the proof of Theorem 3.7 (i), given an infinite word, we can always replace v by u to arrive at a v-free infinite word.Condition 2.3.ensures that two distinct v-free infinite words x and y determine distinguished characters χ x χ y ∈ Y : Let x, y be v-free infinite words with x y, say x ≡ pa • • • and y ≡ pb • • • for some p ∈ {a, b} * .For some i and j, we have x i ≡ [x] i ≡ par and y j ≡ [y] j ≡ pbs for some r, s ∈ {a, b} * with ℓ * (r), ℓ * (s) ≥ L. Now suppose that χ ∈ Ω satisfies χ(x i P) = 1 = χ(y j P).Then χ((parP) ∩ (pbs)P) = 1.Condition 2.3.implies that (parP) ∩ (pbs)P ⊆ pwP ⊆ wP, where the last inclusion is due to Lemma 4.1 (i).Hence χ(wP) = 1, so that χ Y .Therefore, condition 2.3.implies that elements in Y are in one-to-one correspondence to v, w-free infinite words.
The proof that the graphs constructed in cases 1 and 2 are indeed graph models is now similar to the one of Theorem 3.7.
or consider u = a p and b = b q for p, q ≥ 2. The corresponding presentations ({a, b} , u = v) both define Garside monoids, where w = u = v is a Garside element.Thus the conditions in case 1 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 [39] that these are the only Garside groups on two generators (see also [19,Question 1]).r (G ⋉ Ω∞ ) are purely infinite simple.This is straightforward to check using, for instance, [43,Theorem 3.15].(Note that the conventions in [43] and [31] 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, is not unital and hence must be stable by [47].As UCT Kirchberg algebras are classified by their K-theory, it is a natural task to compute K-theory for ).We will discuss K-theory in § 5.4.In order to discuss some examples for case 2, let us formulate a stronger condition which implies condition 2. Example 4.7.Let u = b j and v = ab i ab i a . . .ab i a, with i > 0 and ℓ b (u) > ℓ b (v).Then we saw in § 2.1 that P = a, b | u = v + 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 2 are satisfied.To verify condition 2.3'., note that [v] l σ l+1 is either of the form ab i ab i a . . .ab i+1 (with less letters a appearing as in v) or ab i ab i a . . .ab 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

Artin-Tits monoids of finite type.
Let us now study Artin-Tits monoids of finite type with more than two generators.We will see that the results in § 4.1 do not carry over.Indeed, G ⋉ Ω∞ is typically not amenable.Moreover, our structural results for C * r (G ⋉ Ω∞ ) and thus for the semigroup C*-algebras of Artin-Tits monoids of finite type complement nicely the results in [8,9,22] on C*-algebras of right-angled Artin monoids.
The following refers to [37].Thus G ≡ G S is the group with finite generating set S, and presentation S | (s, t) m s t = (t, s) m s t , for s, t ∈ S , where (s, t) m = ststs... with m factors, and M = (m st ) s,t ∈S is a symmetric positive integer matrix with m ss = 1 and m st > 1 if s t.Moreover, if W is the group with the same presentation as G but with the additional relations {s 2 : s ∈ 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 ) We let P ≡ P S be the monoid given by S | (s, t) m s t = (t, s) m s t for s, t ∈ S + .As in [37] we let p : G → W be the obvious surjection, and P red = {x ∈ P : ℓ(x) = ℓ(p(x))}, where ℓ is the length function.For s, t ∈ S we write ∆ s,t = (s, t) m s t .We use the following notation from [37]: for g, h ∈ P we write g ≺ h if h ∈ gP and h ≻ g if h ∈ Pg.By [37, Proposition 2.6], for any two elements g, h ∈ P there is a unique maximal (for ≺) element g ∧ h ∈ P satisfying g ∧ h ≺ g, h, and as in [37, Section 3] also a unique minimal element g ∨ h ∈ P satisfying g, h ≺ g ∨ h.It follows that P is left reversible and right LCM.For g ∈ P we let L(g) = {s ∈ S : s ≺ g} and R(g) = {s ∈ S : g ≻ s}.For g ∈ P red and s ∈ S, if s R(g) then gs ∈ P red , and similarly on the left (see [37, remarks before Proposition 1.5]).A finite sequence (g 1 , . . ., g k ) of elements of We use the notation ν(g) from [37, p. 367] for the decomposition length of g ∈ P:   As in [37, Section 3], there is a unique element ∆ ∈ P red of greatest length (which plays the role of w in Lemma 4.1).We will write P 0 = P red \ {1, ∆}.Then for 1 g ∈ P \ ∆P, g has normal form (g 1 , . . ., g k ) with g i ∈ 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 Ω given in [44, Section 7].For g ∈ P let [g] = {h ∈ P : h ≺ g}.A subset x ⊆ P is hereditary if [g] ⊆ x whenever g ∈ x, and is directed if for all g, h ∈ x, gP ∩ hP ∩ x ∅.Then Ω can be identified with the collection of all directed hereditary subsets of P. Left reversibility implies that P itself is a directed hereditary subset, corresponding to ∞ ∈ Ω.The finite directed hereditary subsets correspond to the elements of P itself, via g ∈ P ↔ [g] ⊆ P. Thus Ω∞ corresponds to {x ⊆ P : x is directed, hereditary, infinite, and x P}.For k ≥ 0 let X k = {x ∈ Ω∞ : ∆ k ∈ x, ∆ k+1 x}.Then Ω∞ = ⊔ ∞ k=0 X k .Lemma 4.12.X 0 is in one-to-one correspondence with infinite sequences (g 1 , g 2 , . ..) such that g i ∈ P 0 and R(g i ) ⊇ L(g i+1 ) for all i.The correspondence pairs such a sequence with the directed hereditary set Proof.First we claim that if x ∈ X 0 then x contains a unique element of P 0 of maximal length.To see this, first consider two elements g, h ∈ P 0 ∩ x.Since x is directed there is f ∈ x ∩ gP ∩ hP.But then g ∨ h ≺ f .Since x is hereditary, we have g ∨ h ∈ x.Since g, h ∈ P 0 we have g, h ≺ ∆, and hence g ∨ h ∈ P red .But since g ∨ h ∈ x and x ∈ X 0 we know that g ∨ h ∆.Therefore g ∨ h ∈ P 0 ∩ x.Since x ∩ P 0 is finite we may apply the preceding argument finitely many times to see that (x ∩ P 0 ) is the desired element of maximal length in x ∩ P 0 .Now let x ∈ X 0 and let g 1 be the maximal element of x ∩ P 0 .We write σ g 1 (x) = {h ∈ P : g 1 h ∈ x} = g −1 1 (x ∩ g 1 P) (with notation borrowed from [44]).It is easy to check that σ g 1 (x) ∈ X 0 .Let g 2 be maximal in σ g 1 (x) ∩ P 0 .Then R(g 1 ) ⊇ L(g 2 ): for, if s ∈ L(g 2 ) \ R(g 1 ) we may write g 2 = sg ′ 2 , and also we have g 1 s ∈ P red .Since g 1 s ≺ g 1 g 2 we know that g 1 s ∈ x, and hence g 1 s ∈ P 0 , contradicting the maximality of g 1 .Repeating this construction we obtain an infinite sequence (g 1 , g 2 , . ..) such that R(g i ) ⊇ L(g i+1 ) for all i.Since g 1 • • • g n ∈ x for all n, we have Then necessarily g 1 is the maximal element of P red ∩ [ f ].Therefore either h ∈ P red and h ≺ g 1 , or h P red and g 1 ≺ h.In the latter case, g 1 is the maximal element of P red ∩ [h], and hence h has normal form (g 1 , h 2 , • • • , h k ).Repeating this argument for σ g 1 (x) and σ g 1 (h), we find that the normal form of h is (g 1 , . . ., We have shown that for each x ∈ X 0 there is an infinite sequence (g 1 , g 2 , . ..) such that R(g i ) ⊇ L(g i+1 ) for all i and then g 1 = h 1 since both equal the unique maximal element of x ∩ P 0 .Inductively, g i+1 = h i+1 is the unique maximal element of σ g 1 •••g i (x) ∩ P 0 .Therefore the sequence (g 1 , g 2 , . ..) associated to x is unique.Finally, if (g 1 , g 2 , . ..) is any sequence in P 0 such that R(g Remark 4.13.The above proof actually shows that Ω∞ is in one-to-one correspondence with sequences (g 1 , g 2 , . ..) such that g i ∈ P red \ {1}, R(g i ) ⊇ L(g i+1 ) for all i, and g i ∈ P 0 eventually.
In this case Z(g) ∩ Z(h) ∩ X 0 = Z(g ∨ h) ∩ X 0 .The collection {Z(g) \ n j=1 Z(h j ) : h j ∈ gP for 1 ≤ j ≤ n} is a base of compact-open sets for the topology of Ω∞ .Then X 0 is a compact-open subset of Ω∞ that is transversal, and hence the groupoid G ⋉ Ω∞ 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 < ∞.Then G ⋉ Ω∞ is not amenable.
Proof.Let G st be the subgroup generated by s and t.Then G st is nonamenable.Let x = (∆ s,t , ∆ s,t , . ..) ∈ X 0 .Then the isotropy of G ⋉ Ω∞ at x contains G st , and hence is nonamenable.Since there is a point with nonamenable isotropy, the groupoid is nonamenable.
We now give several lemmas on the existence of certain normal forms.Note that since ∆ is the unique element of P red of maximal length, s ∈ L(∆) and s ∈ R(∆) for all s ∈ S. Moreover ∆ is the only element of P red with this property: if g ∈ P red is such that s ≺ g for all s ∈ S, then ℓ(p(s)p(g)) = ℓ(p(g)) − 1 < ℓ(p(g)) for all s ∈ S, so p(g) must have maximal length in W. Therefore g = ∆ (and a similar argument works on the right).If T ⊆ S we write ∆ T for the unique element in P T,red of maximal length (P T,red ⊆ P red by Proposition 4.11).Lemma 4.16.Let T ⊆ S and g ∈ P red be such that T ⊆ L(g).Then ∆ T ≺ g.
Proof.Let M = {h ∈ P : h ∈ T + and h ≺ g}.Since g ∈ P red we know that M ⊆ P red , and so M is finite.We verify the two conditions in [37, Lemma 1.4].For the first condition, let h ∈ M and k ∈ P be such that k ≺ h.By Proposition 4.11 we have k ∈ P T , and hence k ∈ M. For the second condition, let k ∈ P and s, t ∈ S such that k s, kt ∈ M. By the first condition k ∈ M ⊆ P T .Therefore s, t ∈ G T , hence s, t ∈ T. Since k ∈ M we have k ≺ g, so g = kg ′ .Then k s, kt ≺ kg ′ , so s, t ≺ g ′ .By [37, Proposition 1.5] we have ∆ s,t ≺ g ′ , and hence k∆ s,t ≺ g.Therefore k∆ s,t ∈ M. Now by [37,Lemma 1.4] there is f ∈ M such that M = {h ∈ P : h ≺ f }.Then f ∈ P T,red .Since T ⊆ M we have t ≺ f for all t ∈ T. Then t f P T,red for all t ∈ T, so f is maximal in P T,red .Therefore f = ∆ T .
Note that we obtain an alternative proof for Lemma 4.16 by observing that the canonical map P T → P S is ∨preserving.This, in turn, follows for instance from the recipe for computing ∨ using right reversing (see [18, Proposition 6.10]), which does not depend on the ambient monoid.Lemma 4.17.Let T, U ⊆ S be such that m tu = 2 for all t ∈ T and u ∈ U (in particular, T and U are disjoint).Let g ∈ P T .Then L(g∆ U ) = L(g) ∪ U and R(g∆ U ) = R(g) ∪ U.
Proof.The containments ⊇ are clear.Let s ∈ L(g∆ U ) and suppose that s U. Then since U ⊆ L(g∆ U ), Lemma 4.16 implies that ∆ U∪{s } ≺ g∆ U = ∆ U g, hence ∆ U s ≺ ∆ U g, and hence s ≺ g.Therefore s ∈ L(g) as required.A similar argument on the right gives the second statement of the lemma.
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 , . . ., a p ∈ S be such that m a µ ,a µ+1 > 2 for 1 ≤ µ < p and m a µ ,a ν = 2 if | µ − ν| ≥ 2 (so that a 1 , . . ., a p forms a linear subgraph of the Coxeter diagram of G).For 1 ≤ µ ≤ 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 ⊇ hold.We prove the reverse containments.
Proof.If ℓ = 2 we have the normal form (us 1 s 2 , s 2 s 1 s 3 u, us 1 s 3 s 4 , . . .us 1 s k−1 s k ): the L and 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 Lemma 4.34.We consider the situation of Remark 4.30.Let 1 < i < k.Then {u, s 1 , . . ., s i−1 } ∼ {u,s 1 ,...,s i+1 } {u, s 1 , . . ., s i }.
Proof.This is the same as Lemma 4.24 with u included in all L and 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.Lemma 4.35.We consider the situation of Remark 4.30.Let Proof.This is the same as Lemma 4.25 with u included in all L and R sets.The proof is analogous to that of Lemma 4.34.
In the case ℓ = 3 we use the normal form . The L and above we have x ∈ U. Thus Z(ε) ⊆ U. We will write the normal form of ε as (ε 1 , . . ., ε m ).It follows almost immediately that the restriction to X 0 is minimal: if x ∈ X 0 , then εx ∈ Z(ε) ⊆ 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 V ⊆ U, and such that there is b ∈ P with bV V. Since S has cardinality at least three we may choose u ∈ S such that u t and L(ε 1 ) {u}.Then u ∈ R(ε m ).By Proposition 4.38 there is a ∈ P with normal form (a 1 , . . ., a ℓ ) such that L(a 1 ) = {u} and R(a ℓ ) ⊇ L(ε).Then elements of εuaZ(ε) have infinite normal forms (ε 1 , . . ., ε m , u, a 1 , . . ., a ℓ , ε 1 , . . ., ε m−1 , . ..), while elements of dεu 2 aZ(ε) have infinite normal forms (ε 1 , . . ., ε m , u, u, a 1 , . . ., a ℓ , ε 1 , . . ., ε m−1 , . ..).This follows from the preliminary result above, applied to c = εuaε and c = εu 2 aε.It follows that εuaZ(ε) and εu 2 aZ(ε) are disjoint subsets of Z(ε).Therefore letting b = εua we have that bZ(ε) Z(ε).Now we show that the restriction to X 0 is topologically principal.We must find an element z ∈ U with trivial isotropy.Let u ∈ R(y p ), and let t ∈ S with m ut > 2. Let (i 1 , i 2 , . ..) be an aperiodic sequence in ∞ 1 {1, 2}.In the following, for s ∈ S we will write (s) i for the normal form (s, s, . . ., s) with i terms.Let z ∈ X 0 have normal form (y 1 , . . ., y p , u, ut, (t) i 1 , tu, ut, (t) i 2 , tu, ut, . ..).Then z ∈ U, by the first claim at the beginning of the proof of the Theorem.Let d be such that dz ∈ X 0 .Let d have normal form (d 1 , . . ., d k ).Let us consider the product . By [37,Lemma 4.6] we can write Continuing we see that for all large enough r we have that dz 1 • • • z r has normal form (• • • , ut, (t) i k , tu, ut, (t) i k+1 , tu, . ..).There exists a positive integer q with (z q+1 , z q+2 , z q+3 , . ..) = (ut, (t) i k , tu, . ..) and such that for all r ≥ q, the normal form of dz 1 • • • z r is given by the normal form of dz 1 • • • z q followed by z q+1 , z q+2 , z q+3 , . . ., z r .So the normal form of dz is given by the normal form of dz 1 • • • z q followed by ut, (t) i k , tu, ut, (t) i k , tu, . ... Now if dz = z, then thistogether with aperiodicity of (i j ) -implies that the normal form of dz 1 • • • z q must coincide with (z 1 , . . ., z q ), thus Recall that Ω = Ω \ {∞} is G-invariant, so that we obtain a partial action G C 0 ( Ω) ⊗ A by restriction.As G is exact [25], we obtain an exact sequence (8) 0 In case P ⊆ G is Toeplitz, we know how to compute K-theory for (C(Ω) ⊗ A) ⋊ r G because of [13,14].Therefore, our goal now is to deduce a six term exact sequence in K-theory which allows us to compute K-theory for This will then enable us to determine the K-theory of A ⋊ r G.
Let us explain our strategy.Recall that we defined X = { χ ∈ Ω: χ(wP) = 0}, with w ∈ P as in Lemma 4.1.As X is a compact open subspace of Ω meeting every G-orbit, the projection p := (1 X ⊗ 1) (where Note that we have an isomorphism Here P ∩ X = {x ∈ P: x wP}, e x,y is the rank one operator corresponding to the basis vectors δ x and δ y in ℓ 2 (P ∩ 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 § 4, it is possible to compute K-theory for p (C 0 ( Ω∞ ) ⊗ A) ⋊ r G p.
Hence the K-theoretic six term exact sequence provided by ( 9) already provides a way to compute K-theory for p (C 0 ( Ω) ⊗ A) ⋊ r G 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 E, where we twist C * (E) by A using the G-action γ.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 [11].
5.1.Graph algebras twisted by coefficients.Let E be a graph with set of vertices E 0 , set of edges E 1 and range and source maps r, s.We assume that 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 E 1 of E. Let S E be the inverse semigroup attached to E as in [31, § 3.2], and E E its semilattice of idempotents.Set X E := E E .X E can be identified with the set of finite and infinite paths in E. Let W E be the subspace of finite paths and Y E the subspace of infinite paths.Note that in our case, Y E = ∂E.As explained in [31, § 3.2], there is a canonical partial action F X E such that C(X E ) ⋊ r F is canonically isomorphic to the Toeplitz algebra of E and C(Y E ) ⋊ r F is canonically isomorphic to the graph algebra of E.Moreover, we have an exact sequence 0 Now assume that A is a C*-algebra with an F-action γ.Then we can form the diagonal partial action F C(X E )⊗ A. By exactness of F, we obtain an exact sequence (11) 0 Let p v , v ∈ E 0 and s e , e ∈ E 1 , be the canonical generators (projections and partial isometries) of the Toeplitz algebra of E, so that we may consider the elements p v and s e s * e in C(X E ).A combination of [14,Corollary 3.14] and [33, § 6.1] (see also [32]) yields that the homomorphisms 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 In order to compare the exact sequence (11) with (9), we need to establish a universal property for (C(X E ) ⊗ A) ⋊ r F. Let us assume that A is unital and treat the non-unital case later.Let T A be the universal C*-algebra which comes with a homomorphism ι : A → T A such that T A is generated by projections p v , v ∈ E 0 , partial isometries s e , e ∈ E 1 , and ι(A), subject to the relations that p v , v ∈ E 0 , are pairwise orthogonal, s * e s e = p s(e) , p v ≥ e∈r −1 (v) s e s * e and s e ι(a) = ι(γ e (a))s e for all e ∈ E 1 , v ∈ E 0 and a ∈ A. This means that whenever we have a C*-algebra T, a homomorphism i : A → T, projections q v , v ∈ E 0 , and partial isometries t e , e ∈ E 1 , such that q v , t e and i(a) satisfy analogous relations as the generators of T A , then there is a (unique) homomorphism T A → T sending p v to q v , s e to t e and ι(a) to i(a).Hence, by universal property of T A , there is a homomorphism (15) Here 1 v is the characteristic function of the subspace of X E of all finite and infinite paths with range v.We see that ι : A → T A must be injective, because A → (C(X E ) ⊗ A) ⋊ r F, a → 1 ⊗ a is.So we will from now on identify A with ι(A) and write a instead of ι(a).Our goal is to show that the homomorphism in ( 15) is an isomorphism.
For a finite path µ ∈ E * , where First of all, it is easy to see that T A = span s µ s * ν a: a ∈ A, µ, ν ∈ E * , s(µ) = s(ν) .Moreover, by universal property of T A , for every z ∈ T there is an isomorphism δ z : T A T A given by δ z (a) = a, δ z (p v ) = p v and δ z (s e ) = zs e .The following is easy to see (as in the case of ordinary graph algebras without coefficients): For non-unital A, just apply our constructions above to the unitalization A˜and then restrict all the maps to the crossed products with A in place of A˜.
The isomorphism (10) implies that the homomorphism A → p ((C 0 (P) ⊗ A) ⋊ r G) p, a → 1 {e } ⊗ 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 ϕ = v ∈E 0 ϕ v with In particular, for the concrete graph models given in case 1 and case 2 of Theorem 4.3, we see that ϕ is surjective.So there always exist graph models with surjective ϕ.This is the only requirement we need on our graph models later on.
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) . . .
Let us verify exactness at G i : Take z ∈ G i with π i (z) = 0. Then ψ i (p i (z)) = q i (π i (z)) = 0, so that p i (z) = 0.So there exists y ∈ Ǧi with j i (y

Now we show exactness at H
Finally, for exactness at ker (ϕ i+1 ), let z ∈ ker (ϕ i+1 ) satisfy j i+1 (z) = 0. Then there is y ∈ Ḡi with Let us now apply Lemma 5.4 to the diagram induced by ( 16) in K-theory, where E is a graph model for (G ⋉ Ω∞ ) | Y such that the map ϕ in ( 17) is surjective.Let us write I := p (C 0 ( Ω) ⊗ A) ⋊ r G p. Plugging ( 14) and ( 17) into the Now the following map where the double vertical line indicates the position of v. Pre-composing j ′ i by this isomorphism yields the homomorphism j′ i : with ji 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 E 0 as in ( 26), the original homomorphism j i in ( 20) is given by where The map ϕ i in ( 20) is given by ϕ = (y, 0, . . ., 0, ω, z, 0, . . ., 0), where where ω is at the position corresponding to the finite word a p−1 in (26) and z is at the position corresponding to the finite word ba in (26).

Now the following map
where the double vertical line indicates the position of a p−1 .Pre-composing j ′ i by this isomorphism yields the homomorphism j′ i : Finally, by performing elementary column and row operations corresponding to pre-and post-composition with isomorphisms, j′ i is transformed to j′′ i : with ji as in (30).This completes the proof in the torus knot case.
Remark 5.6.The proof shows that ji , πi and ∂i are related to the maps j i , π i and 21) in the following way: We have isomorphisms ker where the lower horizontal arrow is given by * 0 0 ji for some automorphism * : K i K i .Then πi is given by the composition where the first map is the canonical inclusion, and ∂i is given by the composition where the last map is the canonical projection.In particular, we have im ( πi ) = im (π i ).

5.4.
Examples.We will use the following fact several times.
Consider the case I 2 (m) with m ≥ 2 is even.Then (28) gives that j0 is the zero map, and consequently K 1 (I) Z and K 0 (I) Z 2 .Moreover, (22) now gives the exact sequence the zero map (using the formula (23)), and π 0 is an isomorphism, we get that ι 0 is the zero map.Hence, K 0 . It follows from Remark 5.7 that the K 1 -group is generated by [λ(a)] and [λ(b)], while K 0 is generated by [1] and a Bott element associated with λ(a) and λ(b).To see the latter, notice that there is a homomorphism I 2 (m) ։ I 2 (2) = Z 2 with a → a, b → b, and the class of the Bott projections generated by a and b corresponds under the map K 0 (C * r (I 2 (m))) → K 0 (C * r (Z 2 )).Moreover, having computed the maps j i and j ′ i in (31) and (32), it is easy to determine the K-theory of In case m ≥ 4, Remark 4.6 tells us that this determines the stable UCT Kirchberg algebra Q up to isomorphism.Furthermore, we have exact sequences Assume next that m ≥ 3 is odd.Then (29) gives that j0 is the map 2 1 , which is injective with cokernel isomorphic to Z. Therefore, K 0 (I) Z and ∂1 is the zero map, so K 1 (I) = 0. From (22) we get the exact sequence Finally, consider the torus knot group T(p, q) of type (p, q) for p, q ≥ 2, and set g = gcd(p, q).Then (30) gives that j0 is the map p q , which is injective with cokernel isomorphic to Z ⊕ (Z/gZ).Therefore, K 0 (I) Z ⊕ (Z/gZ) and ∂1 is the zero map, so K 1 (I) = 0. From (22) we get the exact sequence 0 −→ K 1 (C * λ (T(p, q))) −→ K 0 (I) ι 0 −→ K 0 (A) −→ K 0 (C * λ (T(p, q))) −→ 0. Since ι 0 • π 0 is the zero map, and π 0 is surjective, then ι 0 is the zero map.Hence, K 0 (C * λ (T(p, q))) K 0 (A) Z and K 1 (C * λ (T(p, q))) K 0 (I) Z ⊕ (Z/gZ).The K 0 -group is generated by the identity, while the K 1 -group is generated by Having computed the maps j i and j ′ i in (33) and (34), it is easy to determine the K-theory of Q := C * r (G ⋉ Ω∞ ): We get K 0 (Q) Z, K 1 (Q) Z if p = q = 2, and K 0 (Q) Z/((p − 1)(q − 1) − 1)Z, K 1 (Q) {0} otherwise.In case (p, q) (2, 2), Remark 4.6 tells us that this determines the stable UCT Kirchberg algebra Q up to isomorphism, and we get Q K ⊗ O (p−1)(q−1) .Thus we obtain exact sequences 0 → J → C * λ (P) → C * λ (T(p, q)) → 0 and 0 → K(ℓ 2 P) → J → K ⊗ O (p−1)(q−1) → 0.Moreover, using the formula (23), we see that ι i • π i : K i (A) ⊕ K i (A) → K i (A) is the zero map for i = 0 and 0 −1 1 −1 1 2 0 1 for i = 1.In both cases, π i is surjective (as im (π i ) = im ( πi ) by Remark 5.6), so ι 0 = 0, and ι 1 maps Z 2 ⊕ (Z/2Z) onto Z 2 .Therefore, (22) gives an exact sequence Example 5.10.Artin's representation of braid groups is defined for n ≥ 3 as the action γ of B n on F n , with canonical generators σ i and x i , respectively, given by We consider the case n = for i = 1.Since π 1 is surjective (im (π 1 ) = im ( π1 ) by Remark 5.6), we compute that for ι 1 : Z 4 → Z 3 that ker ι 1 Z 2 , im ι 1 Z 2 , and coker ι 1 Z.On the other hand, the image of π 0 is Z (as im (π 0 ) = im ( π0 ) by Remark 5.6), so the kernel of ι 0 is either Z or Z 2 .If the former holds, then [1] 0 = 0, which is not possible, because the canonical trace does not vanish on [1] 0 .Thus, we must have that the kernel is Z 2 , which means that ι 0 is the zero map, so we get  (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.R

Remark 2 . 2 .
Note that there are examples of presentations (Σ, u = v) which are not r-homogeneous but where Σ, u = v + is still right LCM.For instance, let Σ = {a, b} and u = b d ab c for some positive integers c, d, and v = a.If we define x i := ab ic for i = 0, 1, . . ., then x i = b d x i+1 for all i.Thus Σ | u = v + is not r-Noetherian in the sense of [18, Definition 2.6], so that (Σ, u = v) cannot be r-homogeneous by [18, Proposition 4.3].However, Σ | u = v + is right LCM by [45, Proposition 2.10].

Example 3 . 9 .
Here is a concrete class of examples where Corollary 3.8 applies: Let A and B be sets with | A| + |B| ≥ 3. Choose u ∈ A * arbitrary and v ∈ B * with OVL(v) = {ε}.Then the presentation (A ∪ B, u = v) satisfies all the assumptions in Theorem 3.7 and Corollary 3.8, so that Corollary 3.8 applies to the monoid P = A ∪ B | u = v + .

1 .
obviously a compact open subspace of Ω ∞ which meets every G-orbit, we have G ⋉ Ω∞ ∼ M (G ⋉ Ω∞ ) | Y .Hence it suffices to describe the restriction (G ⋉ Ω∞ ) | Y of G ⋉ Ω∞ to Y .Let us now -in the same spirit as Theorem 3.7 -present two cases where we can provide a graph model for (G ⋉ Ω∞ ) | Y .Theorem 4.3.Assume that in Lemma 4.1, we may take w = u = v.Let l := max(ℓ * (u), ℓ * (v)) − 1.Then a graph model for (G ⋉ Ω∞ ) | Y is given as follows:

Remark 4 . 8 . 4 . 9 .
As in Remark 4.6, it is straightforward to check that for the presentations in Example 4.7, C * r ((G ⋉ Ω∞ ) | Y ) is a unital UCT Kirchberg algebra and C * r (G ⋉ Ω∞ ) is a stable UCT Kirchberg algebra.Remark There is an overlap between the two classes of groups in Example 4.5, because the presentations ({a, b} , aba • • • = bab • • • ), where each of the relators has m factors, with m odd, and ({a, b} , 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.
we write G T , respectively P T , for the Artin-Tits group, respectively monoid, corresponding to {m st : s, t ∈ T }.It is clear that the inclusion T ⊆ S defines a homomorphism G T → G S carrying P T to P S .Lemma 4.10.Let T ⊆ S and let g ∈ T G S .Then L(g), R(g) ⊆ T.

Proposition 4 . 11 .
Let T ⊆ S. Then the homomorphism G T → G S is injective.Proof.Let g ∈ P T .Then g has a normal form (g 1 , . . ., g k ) in P T .By Lemma 4.10 this is also a normal form in P S .By[37, Corollary 4.3]  it follows that G T → G S is injective on P T .By[37, Corollary 3.2]  it follows that G T → G S is injective.

Remark 4 . 23 .Lemma 4 . 24 .
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.Let T = {s 1 , . . ., s k } ⊆ S be a linear subgraph of the Coxeter diagram of G (as in applies.If ℓ = 3 we have the normal form (us 1 s 2 , us 2 s 1 s 3 , s 1 s 3 s 4 u, us 1 s 4 s 5 , . . ., us 1 s k−1 s k ).The arguments here are analogous.The cases ℓ = 2, k = 3 and ℓ = 3, k = 4 are slightly different.When ℓ = 2 and k = 3 we use the normal form (us 1 s 2 , s 2 s 1 s 3 u).The calculation of L(s 2 s 1 s 3 u) and R(s 2 s 1 s 3 u) are as in the proof of Lemma 4.18.When ℓ = 3 and k = 4 we use the normal form (us 1 s 2 , us 2 s 1 s 3 , s 1 s 3 s 4 u); again, calculation of the relevant L and R sets is as in the proof of Lemma 4.18.

and hence d = 1 .
Therefore z has trivial isotropy.Now, simplicity follows from [41, Proposition II.4.6], and pure infiniteness from [2, Proposition 2.4]. 5. E K-Let P = Σ | u = v + be a right LCM monoid as in § 4, and assume that we are in case 1 or case 2 of Theorem 4.3.Let G = Σ | u = v be the group given by the same presentation as P. Suppose A is a C*-algebra and γ : G A is a G-action on A. The partial action G Ω described in § 2.2 induces a partial action G C(Ω), which in turn gives rise to the diagonal partial action G C(Ω) ⊗ A.
0 0 . . .0 β 0 . . .0 is at the position corresponding to the finite word a p−1 in (26)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

Remark 5 . 7 .
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 ∈ G.The quotient G ab = G/G ′ is called the abelianization of G and we let f ab : G → G ab denote the canonical quotient map.The following result can now be deduced from[36] (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 homomorphismG ab → K 1 (C * r(G)) defined by f ab (g) → [λ(g)] 1 .Example 5.8 (A = C).Let A = C, so of course, the action is trivial, K 0 (C) = Z, and K 1 (C) = 0, and (27) gives the exact sequence 0 and π 0 is surjective, then ι 0 is the zero map.Hence,K 0 (C * λ (I 2 (m))) K 0 (A) Z and thus K 1 (C * λ (I 2 (m))) K 0 (I) Z.The K 0 -group is generated by the identity, and by Remark 5.7, theK 1 -group is generated by [λ(a)] = [λ(b)].As above, using(31) and(32), it is easy to determine the K-theory ofQ := C * r (G ⋉ Ω∞ ): We get K 0 (Q) Z/(m − 2)Z and K 1 (Q) {0}.Remark 4.6 tells us that this determines the stable UCT Kirchberg algebra Q up to isomorphism, and we get Q K ⊗ O m−1 (it is possible to write down an explicit isomorphism).So we have exact sequences 0 → J → C * λ (P) → C * λ (I 2 (m)) → 0 and 0 → K(ℓ 2 P) → J → K ⊗ O m−1 → 0.

Lemma 2.3. Every
Let us write w i := [w] i .Then define χ w ∈ Ω by setting χ w (zP) = 1 if and only if w i ∈ zP for some i.It is easy to see that χ w ∈ Ω ∞ .Conversely, we have χ ∈ Ω ∞ is of the form χ w for some w ∈ Σ ∞ .Proof.Given χ ∈ Ω ∞ , let us enumerate zP ∈ J with χ(zP) = 1, so that {zP: χ(zP) = 1} = {z 1 P, z 2 P, . ..}.Then choose y i ∈ Σ * with y i P = z 1 P ∩ . . .∩ z i P and y i+1 ∈ y i Σ * .Note that all y i P themselves lie in {z 1 P, z 2 P, . ..}.Let l i := ℓ * (y i ).Since χ ∈ Ω ∞ , we must have lim i→∞ l i = ∞.So there is a unique infinite word w in Σ such that [w] l i = y i .We claim that χ = χ w .Indeed, given x ∈ P, we have χ(xP) = 1 if and only if xP ∈ {z 1 P, z 2 P, . ..} if and only if y i ∈ xP for some i if and only if [w] j ∈ xP for some j if and only if χ w (xP) = 1.

Theorem 3.1. Assume
that there is ε z ∈ Σ * 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.Then, for every χ ∈ Ω ∞ , G. χ = Ω ∞ if P is finitely generated, and G. χ = Ω if P is not finitely generated.Moreover, there exists ω ∈ Ω ∞ with trivial stabilizer group, G ω = {e}.
[37,ows from[37, Proposition 2.5] that ∆ {s,t 1 } ≺ g.Then since t 1 s ≺ ∆ s,t 1 we have t 1 s ≺ g, and hence s ≺ t 2 • • • t m .Repeating this process we eventually obtain s ≺ t m , hence s = t m ∈ T, contradicting our supposition.The argument for R(g) is similar.