HNN extensions of quasi-lattice ordered groups and their operator algebras

TL;DR

This paper investigates conditions under which HNN extensions of quasi-lattice ordered groups remain quasi-lattice ordered and amenable, extending the understanding of their operator algebra structures.

Contribution

It provides new criteria for HNN extensions of quasi-lattice ordered groups to preserve quasi-lattice order and amenability.

Findings

Conditions for HNN extensions to be quasi-lattice ordered

HNN extensions inherit amenability from base groups under certain conditions

Application to Baumslag-Solitar groups and their operator algebras

Abstract

The Baumslag-Solitar group is an example of an HNN extension. Spielberg showed that it has a natural positive cone, and that it is then a quasi-lattice ordered group in the sense of Nica. We give conditions for an HNN extension of a quasi-lattice ordered group $(G,P)$ to be quasi-lattice ordered. In that case, if $(G,P)$ is amenable as a quasi-lattice ordered group, then so is the HNN extension.

Equations

$$x\le y\iff x^{-1}y\in P\iff y\in xP.$$

$$x\in P{P}^{-1}$$

$$W_p{W}_p^{\ast }{W}_q{W}_q^{\ast }=\{\begin{array}{ll}{W}_{p\vee q}{W}_{p\vee q}^{\ast }& \text{if p∨q<∞}\\ 0& \text{otherwise.}\end{array}$$

$$W_p^{\ast }{W}_q=\{\begin{array}{ll}{W}_{p^{-1}(p\vee q)}{W}_{q^{-1}(p\vee q)}^{\ast }& \text{if p∨q<∞}\\ 0& \text{otherwise.}\end{array}$$

$$C^{\ast }(G,P)=\overline{\mathrm{span}}\{w_p{w}_q^{\ast }:p,q\in P\}.$$

$$x\in {\mu }^{-1}(k)\cap P\text{ and }x\le \sigma \Rightarrow \sigma =x.$$

$$E:C^{\ast }(G,P)\to \overline{\mathrm{span}}\{w_p{w}_p^{\ast }:p\in P\}$$

$${\delta }_G:C^{\ast }(G)\to C^{\ast }(G){\otimes }_{\min }{C}^{\ast }(G)$$

$$(\delta \otimes \mathrm{id})\circ \delta =(\mathrm{id}\otimes {\delta }_G)\circ \delta .$$

$${\delta }_{\mu }:C^{\ast }(G,P)\to C^{\ast }(G,P){\otimes }_{\min }{C}^{\ast }(K)$$

$$W_p{W}_q=w_p{w}_q\otimes u_{\mu (p)}{u}_{\mu (q)}=w_{pq}\otimes u_{\mu (pq)}=W_{pq}\text{ for all p,q∈P.}$$

$$W_x{W}_x^{\ast }{W}_y{W}_y^{\ast }$$

$$=\{\begin{array}{ll}{w}_{x\vee y}{w}_{x\vee y}^{\ast }\otimes 1& \text{if x∨y<∞}\\ 0\otimes 1& \text{otherwise}\end{array}$$

$$=\{\begin{array}{ll}{w}_{x\vee y}{w}_{x\vee y}^{\ast }\otimes w_{\mu (x\vee y)}{w}_{\mu (x\vee y)}^{\ast }& \text{if x∨y<∞}\\ 0& \text{otherwise}\end{array}$$

$$=W_{x\vee y}{W}_{x\vee y}^{\ast }.$$

$$(({\delta }_{\mu }\otimes \mathrm{id})\circ {\delta }_{\mu })(w_p{w}_q^{\ast })$$

$$={\delta }_{\mu }(w_p{w}_q^{\ast })\otimes \mathrm{id}(u_{\mu (p{q}^{-1})})$$

$$=w_p{w}_q^{\ast }\otimes u_{\mu (p{q}^{-1})}\otimes u_{\mu (p{q}^{-1})}$$

$$=w_p{w}_q^{\ast }\otimes {\delta }_K(u_{\mu (p{q}^{-1})})$$

$$=\mathrm{id}\otimes {\delta }_K(w_p{w}_q^{\ast }\otimes u_{\mu (p{q}^{-1})})$$

$$=((\mathrm{id}\otimes {\delta }_K)\circ {\delta }_{\mu })(w_p{w}_q^{\ast }).$$

$$\pi \otimes ϵ:C^{\ast }(G,P){\otimes }_{\min }{C}^{\ast }(G)\to B(H)\otimes \mathbb{C}=B(H).$$

$$(\pi \otimes ϵ)\circ {\delta }_{\mu }(w_p)=(\pi \otimes ϵ)(w_p\otimes u_{\mu (p)})=\pi (w_p),$$

$${\delta }_{\mu }(C^{\ast }(G,P))(1\otimes C^{\ast }(K))=C^{\ast }(G,P){\otimes }_{\min }{C}^{\ast }(K).$$

$${\delta }_{\mu }(w_p{w}_q^{\ast })(1\otimes u_{\mu (q{p}^{-1})k})=w_p{w}_q^{\ast }\otimes u_{\mu (p{q}^{-1})}(1\otimes u_{\mu (q{p}^{-1})k})=w_p{w}_q^{\ast }\otimes u_k.$$

$$\tau (u_k)=({\lambda }_k{ϵ}_e\mid {ϵ}_e)=\{\begin{array}{ll}1& \text{if k=e}\\ 0& \text{otherwise}.\end{array}$$

$${\delta }_{\mu }:C^{\ast }(G,P)\to C^{\ast }(G,P){\otimes }_{\min }{C}^{\ast }(K)$$

$${\Psi }^{\mu }:=(\mathrm{id}\otimes \tau )\circ {\delta }_{\mu }$$

$${\Psi }^{\mu }(w_p{w}_q^{\ast })=\{\begin{array}{ll}{w}_p{w}_q^{\ast }& \text{if μ(p)=μ(q)}\\ 0& \text{otherwise,}\end{array}$$

$$ϵ$$

$$\displaystyle=\Big{\|}\Psi^{\mu}(a)-\Psi^{\mu}\Big{(}\sum_{(p,q)\in F}\lambda_{p,q}w_{p}w^{*}_{q}\Big{)}\Big{\|}=\Big{\|}b-\sum_{(p,q)\in F\text{, }\mu(p)=\mu(q)}\lambda_{p,q}w_{p}w^{*}_{q}\Big{\|}.$$

$$0$$

$$=(f\otimes \tau )\circ {\delta }_{\mu }(a^{\ast }a)=\tau \circ (f\otimes \mathrm{id})\circ {\delta }_{\mu }(a^{\ast }a).$$

$$g\circ (f\otimes \mathrm{id})\circ {\delta }_{\mu }(a^{\ast }a)=(f\otimes g)\circ {\delta }_{\mu }(a^{\ast }a)=0.$$

$$f_h\otimes f_k(x)=({\pi }_1\otimes {\pi }_2(x)(h\otimes k)\mid h\otimes k).$$

Full text

This research was supported by the Marsden Fund of the Royal Society of New Zealand and by a University of Otago Publishing Bursary

1. Introduction

Since they were introduced by Nica [9], quasi-lattice ordered groups and their $C^{*}$-algebras have generated considerable interest (see, for example, [5],[6]). The amenability of quasi-lattice ordered groups has been a deep subject (see, for example, [3],[4] and [7]). Quasi-lattice ordered groups are also examples of the more recent LCM semigroups [1], [13]. Here we generalise two recent results about the Baumslag-Solitar group.

First, Spielberg proved that the Baumslag-Solitar group is quasi-lattice ordered [11]. The Baumslag-Solitar group is an example of an HNN extension of $\mathbb{Z}$, and hence we wondered if HNN extensions could provide new classes of quasi-lattice ordered groups. Spielberg also showed that a groupoid associated to the Baumslag-Solitar semigroup is amenable [11, Theorem 3.22].

Second, Clark, an Huef and Raeburn examined the phase-transitions of the Toeplitz algebra of the Baumslag-Solitar group [2]. As part of their investigation they proved that the Baumslag-Solitar group is amenable as a quasi-lattice ordered group. The standard way to prove amenability, introduced by Laca and Raeburn [5], is to use a “controlled map”: an order-preserving homomorphism between quasi-lattice ordered groups. They observed that the height map, which counts the number of times the stable letter of the HNN extension appears in a word, is almost a controlled map, and then they adapted the standard proof in [2, Appendix A] to fit.

Our innovation in this paper is a more general definition of a controlled map. We prove in Theorem 3.2 that if $(G,P)$ is a quasi-lattice ordered group and there is a controlled map $\mu$ into an amenable group, and if $\ker\mu$ is an amenable quasi-lattice ordered group, then $G$ is amenable. The motivation for Theorem 3.2 was two-fold. First, if a normal subgroup $N$ of a group $G$ is amenable and $G/N$ is amenable, then $G$ is amenable, and second, Spielberg’s result on amenability of groupoids [12, Proposition 9.3].

In Theorem 4.1 we give conditions under which an HNN extension of a quasi-lattice ordered group is quasi-lattice ordered. This result allows us to construct many new examples of quasi-lattice ordered groups. Finally, we use Theorem 3.2 to prove that an HNN extension of an amenable quasi-lattice ordered group is amenable (Theorem 5.1).

2. Preliminaries

Let $P$ be a subsemigroup of a discrete group $G$ such that $P\cap P^{-1}=\{e\}$. There is a partial order on $G$ defined by

[TABLE]

The order is left-invariant in the sense that $x\leq y$ implies $zx\leq zy$ for all $z\in G$. A partially ordered group $(G,P)$ is quasi-lattice ordered if every finite subset of $G$ with a common upper bound in $P$ has a least common upper bound in $P$ [9, Definition 2.1]. By [3, Lemma 7], $(G,P)$ is quasi-lattice ordered if and only if:

[TABLE]

Let $(G,P)$ be a quasi-lattice ordered group, and let $x,y\in G$. If $x$ and $y$ have a common upper bound in $P$, then their least common upper bound in $P$ is denoted $x\vee y$. We write $x\vee y=\infty$ when $x$ and $y$ have no common upper bound in $P$ and $x\vee y<\infty$ when they have a common upper bound. An isometric representation of $P$ in a $C^{*}$-algebra $A$ is a map $W:P\to A$ such that $W_{e}=1$, $W_{p}$ is an isometry and $W_{p}W_{q}=W_{pq}$ for all $p,q\in P$. We say that $W$ is covariant if

[TABLE]

Equivalently, $W$ is covariant if

[TABLE]

An example of a covariant representation is $T:P\to B(\ell^{2}(P))$ characterised by $T_{p}\epsilon_{x}=\epsilon_{px}$ where $\{\epsilon_{x}:x\in P\}$ is the orthonormal basis of point masses in $\ell^{2}(P)$.

In [9, §§2.4 and 4.1] Nica examined two $C^{*}$-algebras associated to $(G,P)$. The reduced $C^{*}$-algebra $C^{*}_{r}(G,P)$ of $(G,P)$ is the $C^{*}$-subalgebra of $B(\ell^{2}(P))$ generated by $\{T_{p}:p\in P\}$. The universal $C^{*}$-algebra $C^{*}(G,P)$ of $(G,P)$ is generated by a universal covariant representation $w$; it is universal for covariant representations of $P$ in the following sense: for any covariant representation $W:P\to A$ there exists a unital homomorphism $\pi_{W}:C^{*}(G,P)\to A$ such that $\pi_{W}(w_{p})=W_{p}$. It follows from (2.2) that

[TABLE]

Nica defined $(G,P)$ to be amenable if the homomorphism $\pi_{T}:C^{*}(G,P)\to C^{*}_{r}(G,P)$ is faithful [9, §4.2]. He identified an equivalent condition: there exists a conditional expectation $E:C^{*}(G,P)\to\overline{\operatorname{span}}\{w_{p}w^{*}_{p}:p\in P\}$, and $(G,P)$ is amenable if and only if $E$ is faithful (that is, $E(a^{*}a)=0$ implies $a^{*}a=0$ for all $a\in C^{*}(G,P)$). Laca and Raeburn took this second condition as their definition of amenability [5, Definition 3.4].

3. Order-preserving maps and amenability

A key technique, introduced by Laca and Raeburn in [5, Proposition 6.6]111There is an error in the statement of [5, Proposition 6.6]: the final line should read “if $\mathcal{G}$ is amenable then $(G,P)$ is amenable”., is the use of an order-preserving homomorphism between two quasi-lattice ordered groups which preserves the least upper bound structure. Crisp and Laca called such a homomorphism a controlled map [4]. If $(G,P)$ and $(K,Q)$ are quasi-lattice ordered groups, $\mu:G\to K$ is a controlled map and $K$ is an amenable group, then $(G,P)$ is amenable as a quasi-lattice ordered group by [5, Proposition 6.6]. Motivated by work in [2, Appendix A] we now give a weaker definition for a controlled map. We then follow the program of [2] to generalise [5, Proposition 6.6]. We state this generalisation in Theorem 3.2 below; its proof will take up the remainder of this section.

Definition 3.1**.**

Let $(G,P)$ and $(K,Q)$ be quasi-lattice ordered groups. Let $\mu:G\to K$ be an order-preserving group homomorphism. For each $k\in Q$, let $\Sigma_{k}$ be the set of $\sigma\in\mu^{-1}(k)\cap P$ which are minimal in the sense that

[TABLE]

We say $\mu$ is a controlled map if it has the following properties:

1. (1)

For all $x,y\in G$ such that $x\vee y<\infty$ we have $\mu(x)\vee\mu(y)=\mu(x\vee y)$. 2. (2)

For all $k\in Q$, $\Sigma_{k}$ is complete in the following sense: for every $x\in\mu^{-1}(k)\cap P$ there exists $\sigma\in\Sigma_{k}$ such that $\sigma\leq x$. 3. (3)

For all $k\in Q$ and $\sigma,\tau\in\Sigma_{k}$ we have $\sigma\vee\tau<\infty\Rightarrow\sigma=\tau$.

Theorem 3.2**.**

Let $(G,P)$ and $(K,Q)$ be quasi-lattice ordered groups. Suppose that $\mu:G\to K$ is a controlled map. If $K$ is an amenable group and \big{(}\mu^{-1}(e),\mu^{-1}(e)\cap P\big{)} is an amenable quasi-lattice ordered group, then $(G,P)$ is amenable.

We start by showing that the kernel of a controlled map is a quasi-lattice ordered group.

Lemma 3.3**.**

Let $(G,P)$ and $(K,Q)$ be quasi-lattice ordered groups, and suppose that $\mu:G\to K$ is a controlled map. Then \big{(}\mu^{-1}(e),\mu^{-1}(e)\cap P\big{)} is a quasi-lattice ordered group.

Proof.

It is clear that $\mu^{-1}(e)$ is a subgroup of $G$ and that$\mu^{-1}(e)\cap P$ is a unital semigroup. Suppose that $x,y\in\mu^{-1}(e)$ have a common upper bound $z\in\mu^{-1}(e)\cap P$. We know that $z$ is a common upper bound for $x,y$ in $P$, and hence $x\vee y$ exists in $P$ and $x\vee y\leq z$. Now $\mu(x\vee y)=\mu(x)\vee\mu(y)=e$, and hence $x\vee y\in\mu^{-1}(e)\cap P$. Thus \big{(}\mu^{-1}(e),\mu^{-1}(e)\cap P\big{)} is a quasi-lattice ordered group. ∎

To prove Theorem 3.2 we will show that the conditional expectation

[TABLE]

is faithful. We will use the amenability of $K$ to construct a faithful conditional expectation $\Psi^{\mu}:C^{*}(G,P)\to\overline{\operatorname{span}}\{w_{p}w^{*}_{q}:\mu(p)=\mu(q)\}$, and then show that $E$ is faithful when restricted to $\operatorname{range}\Psi^{\mu}$. To construct $\Psi^{\mu}$ we follow the method of [5, Lemma 6.5] which uses a coaction.

Let $G$ be a discrete group and let $A$ be a unital $C^{*}$-algebra. Let

[TABLE]

be the comultiplication of $G$ which is characterised by $\delta_{G}(u_{g})=u_{g}\otimes u_{g}$ for $g\in G$. A coaction of $G$ on $A$ is a unital homomorphism $\delta:A\to A\otimes_{\min}C^{*}(G)$ such that

[TABLE]

We say that $\delta$ is nondegenerate if $\delta(A)(1\otimes C^{*}(G))=A\otimes_{\min}C^{*}(G)$.

Lemma 3.4**.**

Let $(G,P)$ be a quasi-lattice ordered group. Suppose that there exists a group $K$ and a homomorphism $\mu:G\to K$. Then there exists an injective coaction

[TABLE]

characterised by $\delta_{\mu}(w_{p})=w_{p}\otimes u_{\mu(p)}$ for all $p\in P$.

Proof.

Let $W:P\to C^{*}(G,P)\otimes_{\min}C^{*}(K)$ be characterised by $W_{p}=w_{p}\otimes u_{\mu(p)}$. We will show that $W$ is a covariant representation, and then take $\delta_{\mu}:=\pi_{W}$. Unitaries are isometries and hence $W_{p}$ is isometric for all $p\in P$. Observe that $W_{e}=w_{e}\otimes u_{\mu(e)}=1\otimes 1$, and

[TABLE]

Thus $W$ is an isometric representation. To prove $W$ is covariant, we fix $x,y\in P$ and compute:

[TABLE]

Thus $W$ is a covariant representation of $P$. By the universal property of $C^{*}(G,P)$, there exists a homomorphism $\delta_{\mu}:=\pi_{W}$, which has the desired properties. Since $W_{e}=1\otimes 1$ it follows that $\delta_{\mu}$ is unital.

To prove the comultiplication identity, we compute on generators: for $p,q\in P$ we have

[TABLE]

Hence $(\delta_{\mu}\otimes{\operatorname{id}})\circ\delta_{\mu}=({\operatorname{id}}\otimes\delta_{K})\circ\delta_{\mu}$. Thus $\delta_{\mu}$ is a coaction.

To show that $\delta_{\mu}$ is injective, let $\pi:C^{*}(G,P)\to B(H)$ be a faithful representation. We will show that $\pi$ can be written as a composition of $\delta_{\mu}$ and another representation. Let $\epsilon:C^{*}(K)\to\mathbb{C}$ be the trivial representation on $\mathbb{C}$ such that $\epsilon(u_{k})=1$ for all $k\in K$. By the properties of the minimal tensor product (see [10, Proposition B.13]) there exists a homomorphism

[TABLE]

Since

[TABLE]

we have $\pi=(\pi\otimes\epsilon)\circ\delta_{\mu}$. Now suppose that $\delta_{\mu}(a)=0$ for some $a\in C^{*}(G,P)$. Then $0=(\pi\otimes\epsilon)\circ\delta_{\mu}(a)=\pi(a)$. Since $\pi$ is faithful, $a=0$. Hence $\delta_{\mu}$ is injective.

To prove that $\delta_{\mu}$ is a nondegenerate coaction we must show that

[TABLE]

It suffices to show that we can get the spanning elements $w_{p}w^{*}_{q}\otimes u_{k}$, and this is easy:

[TABLE]

Thus $\delta_{\mu}$ is nondegenerate. ∎

Let $\lambda$ be the left-regular representation of a discrete group $K$. There is a trace $\tau$ on $C^{*}(K)$ characterised by

[TABLE]

It is well-known that if $K$ is an amenable group, then $\tau$ is faithful.

Lemma 3.5**.**

Let $(G,P)$ be a quasi-lattice ordered group. Suppose that there exist a group $K$ and a homomorphism $\mu:G\to K$. Let

[TABLE]

be the coaction of Lemma 3.4. Then

[TABLE]

is a conditional expectation of $C^{*}(G,P)$ with range $\overline{\operatorname{span}}\{w_{p}w_{q}^{*}:\mu(p)=\mu(q)\}$. If $K$ is an amenable group, then $\Psi^{\mu}$ is faithful.

Proof.

Since ${\operatorname{id}}\otimes\tau$ and $\delta_{\mu}$ are linear and norm decreasing, so is $\Psi^{\mu}$. Since $\Psi^{\mu}(w_{e})=1$ the norm of $\Psi^{\mu}$ is $1$. We have

[TABLE]

and hence $\Psi^{\mu}\circ\Psi^{\mu}=\Psi^{\mu}$. Thus $\Psi^{\mu}$ is a conditional expectation by [14].

From (3.1) we see that $\overline{\operatorname{span}}\{w_{p}w^{*}_{q}:\mu(q)=\mu(p)\}\subseteq\operatorname{range}\Psi^{\mu}$. To show the reverse inclusion, fix $b\in\operatorname{range}\Psi^{\mu}$, say $b=\Psi^{\mu}(a)$ for some $a\in C^{*}(G,P)$. Also fix $\epsilon>0$. There exists a finite subset $F\subseteq P\times P$ such that $\|a-\sum_{(p,q)\in F}\lambda_{p,q}w_{p}w^{*}_{q}\|<\epsilon$. Since $\Psi^{\mu}$ is linear and norm-decreasing,

[TABLE]

Thus $b\in\overline{\operatorname{span}}\{w_{p}w^{*}_{q}:\mu(q)=\mu(p)\}$, and $\operatorname{range}\Psi^{\mu}=\overline{\operatorname{span}}\{w_{p}w^{*}_{q}:\mu(q)=\mu(p)\}$.

Now suppose that $K$ is amenable. To see that $\Psi^{\mu}$ is faithful, we follow the proof of [5, Lemma 6.5]. Let $a\in C^{*}(G,P)$ and suppose that $\Psi^{\mu}(a^{*}a)=0$. Let $f$ be an arbitrary state on $C^{*}(G,P)$. Then

[TABLE]

Since $K$ is amenable, $\tau$ is faithful. Hence $(f\otimes{\operatorname{id}})\circ\delta(a^{*}a)=0$. This implies that for all states $f$ on $C^{*}(G,P)$ and states $g$ on $C^{*}(K)$,

[TABLE]

To see that $\delta_{\mu}(a^{*}a)=0$, let $\pi_{1}:C^{*}(G,P)\to H_{1}$ and $\pi_{2}:C^{*}(K)\to H_{2}$ be faithful representations. Then $\pi_{1}\otimes\pi_{2}$ is a faithful representation of $C^{*}(G,P)\otimes_{\min}C^{*}(K)$ on $B(H_{1}\otimes H_{2})$ by [10, Corollary B.11]. Fix unit vectors $h\in H_{1}$, $k\in H_{2}$. There exists a state $f_{h}\otimes f_{k}$ on $C^{*}(G,P)\otimes_{\min}C^{*}(K)$ defined by

[TABLE]

Since $(f\otimes g)\circ\delta_{\mu}(a^{*}a)=0$ for all states $f$ of $C^{*}(G,P)$ and $g$ of $C^{*}(K)$, we have

[TABLE]

Hence $\pi_{1}\otimes\pi_{2}(\delta_{\mu}(a^{*}a))=0$. Since $\pi_{1}\otimes\pi_{2}$ is faithful, $\delta_{\mu}(a^{*}a))=0$. But $\delta_{\mu}$ is injective, and hence $a=0$, and $\Psi^{\mu}$ is faithful. ∎

Next we investigate the structure of

[TABLE]

In the statement of the next lemma, note that we can view C^{*}\big{(}\mu^{-1}(e),\mu^{-1}(e)\cap P\big{)} as a $C^{*}$-subalgebra of $C^{*}(G,P)$.

Lemma 3.6**.**

Let $(G,P)$ and $(K,Q)$ be quasi-lattice ordered groups, and suppose that $\mu:G\to K$ a controlled map. Let $k\in Q$, and let $F$ be a finite subset of $\Sigma_{k}$. Let

[TABLE]

Then $B_{k,F}$ is a closed $C^{*}$-subalgebra of $\overline{\operatorname{span}}\{w_{p}w_{q}^{*}:\mu(p)=\mu(q)\}$.

Proof.

To see that $B_{k,F}$ is contained in $\overline{\operatorname{span}}\{w_{p}w_{q}^{*}:\mu(p)=\mu(q)\}$, it suffices to consider $D$ of the form $w_{\alpha}w_{\beta}^{*}$ where $\mu(\alpha)=\mu(\beta)=e$. Then $w_{\sigma}Dw^{*}_{\tau}=w_{\sigma\alpha}w_{\tau\beta}^{*}$ and $\mu(\sigma\alpha)=\mu(\sigma)=k=\mu(\tau)=\mu(\tau\beta)$. Thus $B_{k,F}$ is a subalgebra of $\overline{\operatorname{span}}\{w_{p}w_{q}^{*}:\mu(p)=\mu(q)\}$.

We will prove the lemma by showing that $B_{k,F}$ is isomorphic to

[TABLE]

By Definition 3.1(3), the elements of $F$ have no common upper bound unless they are equal. So

[TABLE]

Thus $\{w_{\sigma}w^{*}_{\tau}:\sigma,\tau\in F\}$ is a set of matrix units in the $C^{*}$-algebra $\overline{B_{k,F}}$. This gives a homomorphism $\theta:M_{F}(\mathbb{C})\to\overline{B_{k,F}}$ which maps the matrix units $\{E_{\sigma,\tau}:\sigma,\tau\in F\}$ in $M_{F}(\mathbb{C})$ to $\{w_{\sigma}w^{*}_{\tau}:\sigma,\tau\in F\}\subset{B_{k,F}}$. It is easy to check that the formula

[TABLE]

gives a homomorphism \psi:C^{*}\big{(}\mu^{-1}(e),\mu^{-1}(e)\cap P\big{)}\to B_{k,F}. We have

[TABLE]

Each $A\in M_{F}(\mathbb{C})$ is a linear combination of the $E_{\sigma,\tau}$, and hence $\psi(D)\theta(A)=\theta(A)\psi(D)$ for all $A\in M_{F}(\mathbb{C})$ and D\in C^{*}\big{(}\mu^{-1}(e),\mu^{-1}(e)\cap P\big{)}. Since the ranges of $\theta$ and $\psi$ commute, the universal property of the maximal tensor product gives a homomorphism $\theta\otimes_{\max}\psi$ of M_{F}(\mathbb{C})\otimes_{\max}C^{*}\big{(}\mu^{-1}(e),\mu^{-1}(e)\cap P\big{)} into $\overline{B_{k,F}}$.

By [10, Theorem B.18]

[TABLE]

with no closure. So the range of $\theta\otimes_{\max}\psi$ is spanned by $\theta(E_{\sigma,\tau})\psi(D)=w_{\sigma}Dw^{*}_{\tau}$ and hence is $B_{k,F}$. Thus $B_{k,F}$ is a closed $C^{*}$-subalgebra of $\overline{\operatorname{span}}\{w_{p}w_{q}^{*}:\mu(p)=\mu(q)\}$. ∎

Let $\{\epsilon_{x}:x\in P\}$ be the usual basis for $\ell^{2}(P)$. Let $T$ be the covariant representation of $(G,P)$ on $\ell^{2}(P)$ such that $T_{p}\epsilon_{x}=\epsilon_{px}$, and let $\pi_{T}$ be the corresponding homomorphism of $C^{*}(G,P)$ onto $C^{*}_{r}(G,P)$ such that $\pi_{T}(w_{p})=T_{p}$. For $k\in Q$ we consider the subspaces

[TABLE]

Lemma 3.7**.**

Let $(G,P)$ and $(K,Q)$ be quasi-lattice ordered groups, and suppose that $\mu:G\to K$ is a controlled map. Let $k\in Q$. Then

1. (1)

$H_{k}$* is invariant for $\pi_{T}|_{B_{k,F}}$;* 2. (2)

if \big{(}\mu^{-1}(e),\mu^{-1}(e)\cap P\big{)} is amenable, then $\pi_{T}|_{B_{k,F}}$ is isometric on $H_{k}$.

Proof.

For (1), let $\sigma,\tau\in F$ and let $x,y\in\mu^{-1}(e)$ and let $\epsilon_{\gamma z}\in H_{k}$. Then $w_{\sigma}w_{x}w^{*}_{y}w^{*}_{\tau}$ is a spanning element of $B_{k,F}$. Since $\mu(\tau)=k=\mu(\gamma)$ we have

[TABLE]

If $\pi_{T}(w_{\sigma}w_{x}w^{*}_{y}w^{*}_{\tau})\epsilon_{\gamma z}=0$ we are done. Otherwise, to see that $\epsilon_{\sigma xy^{-1}z}$ is back in $H_{k}$, suppose that $y\leq z$. Then $y^{-1}z\in P$. Since $\sigma x\in\mu^{-1}(k)\cap P$ we have $\epsilon_{(\sigma x)(y^{-1}z)}\in H_{k}$. It follows that $H_{k}$ is invariant for $\pi_{T}|_{B_{k,F}}$.

For (2) suppose that \big{(}\mu^{-1}(e),\mu^{-1}(e)\cap P\big{)} is amenable. We will show that $\pi_{T}|_{B_{k,F}}$ is faithful on $H_{k}$. Take $B=\sum_{\sigma,\tau\in F}w_{\sigma}D_{\sigma,\tau}w^{*}_{\tau}\in B_{k,F}$ and suppose that $\pi_{T}(B)|_{H_{k}}=0$. Fix $\gamma,\delta\in F$. Then

[TABLE]

Since $T_{\delta}$ is an injection from $H_{e}$ to $H_{k}$ and $\pi_{T}(B)|_{H_{k}}=0$, it follows that $\pi_{T}(B)T_{\delta}|_{H_{e}}=0$. Thus $\pi_{T}(D_{\gamma,\delta})|_{H_{e}}=0$.

But the restriction

[TABLE]

is the Toeplitz representation of \big{(}\mu^{-1}(e),\mu^{-1}(e)\cap P\big{)}, and hence is faithful by amenability. Thus $D_{\gamma,\delta}=0$. Repeating the argument finitely many times shows that all the $D_{\sigma,\tau}=0$ and hence that $B=0$. Thus $\pi_{T}|_{B_{k,F}}$ is faithful on $H_{k}$, and therefore is isometric. ∎

Lemma 3.8**.**

Let $(G,P)$ and $(K,Q)$ be quasi-lattice ordered groups, and suppose that $\mu:G\to K$ is a controlled map. Let $B_{k}=\overline{\operatorname{span}}\{w_{p}w^{*}_{q}:\mu(p)=\mu(q)=k\}$. Let $\mathcal{F}$ be the set of all finite sets $F\subseteq\Sigma_{k}$. Then $B_{k}=\overline{\cup_{F\in\mathcal{F}}B_{k,F}}$. If \big{(}\mu^{-1}(e),\mu^{-1}(e)\cap P\big{)} is amenable, then $\pi_{T}|_{B_{k}}$ is isometric on $H_{k}$.

Proof.

Observe that $\mathcal{F}$ is a directed set partially ordered by inclusion with $E,F\in\mathcal{F}$ majorismajorizeded by $E\cup F$. If $E\subseteq F$, then $B_{k,E}\subseteq B_{k,F}$. Thus $\{B_{k,F}:F\in\mathcal{F}\}$ is an inductive system with limit $\overline{\cup_{F\in\mathcal{F}}B_{k,F}}$.

For each $F\in\mathcal{F}$ we have $B_{k,F}\subseteq B_{k}$, and $B_{k}$ is closed. Therefore $\overline{\cup_{F\in\mathcal{F}}B_{k,F}}\subseteq B_{k}$. To prove the reverse inclusion it suffices to show that the spanning elements of $B_{k}$ are in $B_{k,F}$ for some $F$. Fix $p,q\in P$ such that $\mu(p)=\mu(q)=k$ and consider $w_{p}w^{*}_{q}$. By Definition 3.1(2), the set $\Sigma_{k}$ of minimal elements is complete, and there exists $\sigma,\tau\in\Sigma_{k}$ such that $\sigma\leq p$ and $\tau\leq q$. Hence there exists $x,y\in P$ such that $p=\sigma x$ and $q=\tau y$. Thus $w_{p}w^{*}_{q}=w_{\sigma x}w^{*}_{\tau y}=w_{\sigma}(w_{x}w^{*}_{y})w^{*}_{\tau}$ and w_{x}w^{*}_{y}\in C^{*}\big{(}\mu^{-1}(e),\mu^{-1}(e)\cap P\big{)}. Since $\{\sigma,\tau\}\in\mathcal{F}$ we have $w_{p}w^{*}_{q}\in B_{k,\{\sigma,\tau\}}$. Thus $B_{k}\subseteq\overline{\cup_{F\in\mathcal{F}}B_{k,F}}$, and equality follows.

Finally, suppose that \big{(}\mu^{-1}(e),\mu^{-1}(e)\cap P\big{)} is amenable. Then $\pi_{T}|_{B_{k,F}}$ is isometric on $H_{k}$ for all $F\in\mathcal{F}$ by Lemma 3.7(2). Since $\pi_{T}$ is isometric on every $B_{k,F}$, its extension to the closure is also isometric. ∎

Let $\mathcal{I}$ be the set of all finite sets $I\subset Q$ that are closed under $\vee$ in the sense that $s,t\in I$ and $s\vee t<\infty$ implies that $s\vee t\in I$.

Lemma 3.9**.**

Let $(G,P)$ and $(K,Q)$ be quasi-lattice ordered groups, and suppose that $\mu:G\to K$ is a controlled map. For each $I\in\mathcal{I}$ let

[TABLE]

Then $C_{I}$ is a $C^{*}$-subalgebra of $\overline{\operatorname{span}}\{w_{p}w_{q}^{*}:\mu(p)=\mu(q)\}$, $C_{I}=\operatorname{span}\{B_{k}:k\in I\}$ and $\overline{\operatorname{span}}\{w_{p}w_{q}^{*}:\mu(p)=\mu(q)\}=\overline{\cup_{I\in\mathcal{I}}C_{I}}$.

Proof.

Fix $I\in\mathcal{I}$. To see that $C_{I}$ is a $C^{*}$-subalgebra, it suffices to show that $\operatorname{span}\{w_{p}w^{*}_{q}:\mu(p)=\mu(q)\in I\}$ is a $*$-subalgebra. It’s clearly closed under taking adjoints. Let $p,q,r,s\in P$ such that $\mu(p)=\mu(q)\in I$ and $\mu(r)=\mu(s)\in I$. Then

[TABLE]

If $w_{p}w^{*}_{q}w_{r}w^{*}_{s}=0$ we are done. So suppose that $w_{p}w^{*}_{q}w_{r}w^{*}_{s}\neq 0$. Then $q\vee r<\infty$. Since $\mu$ is a controlled map and $\mu(p)=\mu(q)$, by Definition 3.1(1),

[TABLE]

Similarly, $\mu(sr^{-1}(q\vee r))=\mu(q)\vee\mu(r)$. Since $I$ is closed under $\vee$ we have $\mu(q)\vee\mu(r)\in I$, and hence $w_{p}w^{*}_{q}w_{r}w^{*}_{s}\in\operatorname{span}\{w_{p}w^{*}_{q}:\mu(p)=\mu(q)\in I\}$. It follows that $C_{I}$ is a $C^{*}$-subalgebra.

For each $k\in I$, we have $B_{k}\subseteq C_{I}$, and so $\operatorname{span}\{B_{k}:k\in I\}\subseteq C_{I}$. To show the reverse inclusion observe that for $w_{p}w^{*}_{q}\in C_{I}$ we have $w_{p}w^{*}_{q}\in B_{\mu(p)}$. Since the finite span of closed subalgebras is closed, $\overline{\operatorname{span}}\{w_{p}w^{*}_{q}:\mu(p)=\mu(q)\in I\}\subseteq\operatorname{span}\{B_{k}:k\in I\}$. Thus $C_{I}=\operatorname{span}\{B_{k}:k\in I\}$. ∎

Proposition 3.10**.**

Let $(G,P)$ and $(K,Q)$ be quasi-lattice ordered groups, and suppose that $\mu:G\to K$ is a controlled map. If \big{(}\mu^{-1}(e),\mu^{-1}(e)\cap P\big{)} is amenable, then $\pi_{T}$ is faithful on $\overline{\operatorname{span}}\{w_{p}w^{*}_{q}:\mu(p)=\mu(q)\}$.

Proof.

By Lemma 3.9, $\overline{\operatorname{span}}\{w_{p}w^{*}_{q}:\mu(p)=\mu(q)\}=\overline{\cup_{I\in\mathcal{I}}C_{I}}$. Thus it suffices to show that $\pi_{T}$ is isometric on each $C_{I}$. Fix $I\in\mathcal{I}$. Suppose that $\pi_{T}(R)=0$ for some $R\in C_{I}$. Then there exist $R_{k}\in B_{k}$ such that $R=\sum_{k\in I}R_{k}$ and then $\sum_{k\in I}\pi_{T}(R_{k})=0$.

We claim that if $k\not\leq j$, then $\pi_{T}(B_{k})|_{H_{j}}=0$ (it then follows that $\pi_{T}(R_{k})|_{H_{j}}=0$). To prove the claim, it suffices to show that $\pi_{T}(w_{q}^{*})\epsilon_{\gamma z}=0$ for all $q\in\mu^{-1}(k)\cap P$ and $\epsilon_{\gamma z}\in H_{k}$. We have

[TABLE]

But $q\leq\gamma z$ implies $k=\mu(q)\leq\mu(\gamma z)=\mu(\gamma)=j$. So $k\not\leq j$ implies $\pi_{T}(w_{q}^{*})\epsilon_{\gamma z}=0$. Hence $\pi_{T}(B_{k})|_{H_{j}}=0$ if $k\not\leq j$ as claimed.

Let $l$ be a minimal element of $I$ in the sense that $x\leq l$ implies $x=l$. Then for $k\in I$, we have $k\not\leq l$ unless $k=l$. Now

[TABLE]

Since \big{(}\mu^{-1}(e),\mu^{-1}(e)\cap P\big{)} is amenable, $\pi_{T}|_{B_{l}}$ is isometric on $H_{l}$ by Lemma 3.8. Thus $R_{l}=0$.

Let $l_{2}$ be a minimal element of $I\backslash\{l\}$. Then we can repeat the above argument to get $R_{l_{2}}=0$. Since $I$ is finite, we can continue to conclude that $R=0$. ∎

We can now prove Theorem 3.2

Proof of Theorem 3.2.

Suppose that $K$ is an amenable group. To see $(G,P)$ is amenable, we will show that the conditional expectation

[TABLE]

is faithful. Let $\Psi^{\mu}$ be the conditional expectation of Lemma 3.5. We have

[TABLE]

and hence $E=E\circ\Psi^{\mu}$.

Since $K$ is an amenable group, $\Psi^{\mu}$ is faithful by Lemma 3.5. Let $P_{z}\in B(\ell^{2}(P))$ be the orthogonal projection onto $\operatorname{span}\{\epsilon_{z}\}$. It is straightforward to show that the diagonal map $\Delta:B(\ell^{2}(P))\to B(\ell^{2}(P))$ given by

[TABLE]

is a conditional expectation such that $\Delta\circ\pi_{T}=\pi_{T}\circ E$ and is faithful.

Now suppose that $R\in C^{*}(G,P)$ and $E(R^{*}R)=0$. Then $E(\Psi^{\mu}(R^{*}R))=0$ and so $\pi_{T}\circ E(\Psi^{\mu}(R^{*}R))=0$. This gives $\Delta\circ\pi_{T}(\Psi^{\mu}(R^{*}R))=0$. Since $\Delta$ is faithful, it follows that $\pi_{T}(\Psi^{\mu}(R^{*}R))=0$. Since \big{(}\mu^{-1}(e),\mu^{-1}(e)\cap P\big{)} is amenable, Lemma 3.10 implies that $\pi_{T}$ is faithful on $\overline{\operatorname{span}}\{w_{p}w_{q}^{*}:\mu(p)=\mu(q)\}=\operatorname{range}\Psi^{\mu}$. Thus $\Psi^{\mu}(R^{*}R)=0$, and then $R=0$ since $\Psi^{\mu}$ is faithful. Hence $E$ is faithful and $(G,P)$ is amenable. ∎

4. Quasi-lattice ordered HNN extensions

Let $G$ be a group, let $A$ and $B$ be subgroups of $G$, and let $\phi:A\to B$ be an isomorphism. The group with presentation

[TABLE]

is the HNN extension of G with respect to $A,B$ and $\phi$. For every HNN extension $G^{*}$ the height map is the homomorphism $\theta:G^{*}\to\mathbb{Z}$ such that $\theta(g)=0$ for all $g\in G$ and $\theta(t)=1$.

Example**.**

Let $c,d\in\mathbb{Z}$. The Baumslag-Solitar group

[TABLE]

is an HNN extension of $\mathbb{Z}$ with respect to $A=d\mathbb{Z}$, $B=c\mathbb{Z}$ and $\phi:A\to B$ given by $\phi(dn)=cn$ for all $n\in\mathbb{Z}$. Then $\mathbb{Z}^{*}$ satisfies the relation $t^{-1}dt=\phi(d)=c$. Let $\operatorname{BS}(c,d)^{+}$ be the subsemigroup of $\operatorname{BS}(c,d)$ generated by $x$ and $t$. Spielberg showed in [11, Theorem 2.11] that $(\operatorname{BS}(c,d),\operatorname{BS}(c,d)^{+})$ is quasi-lattice ordered for all $c,d>0$; he also proved in [11, Lemma 2.12] that $(\operatorname{BS}(c,-d),\operatorname{BS}(c,-d)^{+})$ is not quasi-lattice ordered unless $c=1$.

To work with an HNN extension we use a normal form for its elements from [8, Theorem 2.1]. We choose $X$ to be a complete set of left coset representatives for $G/A$ such that $xA\neq x^{\prime}A$ for $x\neq x^{\prime}\in X$. Similarly, choose a complete set $Y$ of left coset representatives for $G/B$. Then a (right) normal form relative to $X$ and $Y$ of $g\in G$ is a product

[TABLE]

where:

1. (1)

$g_{n}$ is an arbitrary element of $G$. 2. (2)

If $\epsilon_{i}=1$, then $g_{i-1}$ is an element of $X$ 3. (3)

If $\epsilon_{i}=-1$, then $g_{i-1}$ is an element of $Y$.

By [8, Theorem 2.1], for every choice of complete left coset representatives $X$ and $Y$, each $g\in G^{*}$ has a unique normal form.

Our goal is to generalise the properties of the Baumslag-Solitar group with $c,d>0$ to construct quasi-lattice ordered HNN extensions of other quasi-lattice ordered groups.

Let $(G,P)$ be a quasi-lattice ordered group. Let $G^{*}$ be the HNN extension of $G$ with respect to subgroups $A$ and $B$ with an isomorphism $\phi:A\to B$. Let $P^{*}$ be the subsemigroup of $G^{*}$ generated by $P$ and $t$. In general, $(G^{*},P^{*})$ is not a quasi-lattice ordered group. For example, if $c>1$, then $(\operatorname{BS}(c,-d),\operatorname{BS}(c,-d)^{+})$ is not quasi-lattice ordered by [11, Lemma 2.12]. We need some conditions on our subgroups $A$ and $B$ and on the isomorphism $\phi$ which ensure that $(G^{*},P^{*})$ is quasi-lattice ordered.

There are two reasons why $(\operatorname{BS}(c,d),\operatorname{BS}(c,d)^{+})$ is easy to work with. The first is that there are natural choices of coset representatives: $\{0,\ldots,d-1\}$ for $A=d\mathbb{Z}$ and $\{0,\ldots,c-1\}$ for $B=c\mathbb{Z}$. The second is that the subgroup isomorphism $\phi:md\mapsto mc$ takes positive elements to positive elements. In particular, using this choice of representatives, every element $\omega\in\operatorname{BS}(c,d)^{+}$ has a unique normal form

[TABLE]

where $0\leq m_{i}<d$ for $i<n$ and $m_{n}\in\mathbb{N}$. This choice of coset representatives is associated to the division algorithm on $\mathbb{N}$: for every $n\in\mathbb{N}$ we can uniquely write $n=md+r$ for some $m\in\mathbb{N}$ and $0\leq r\leq d-1$.

In general, for $G^{*}$ we would like a natural choice of coset representatives for $G/A$ and $G/B$ so that every element of $P^{*}$ has a unique normal form that is a sequence of elements in $P$ and $t$.

Theorem 4.1**.**

Let $(G,P)$ be a quasi-lattice ordered group with subgroups $A$ and $B$. Suppose that:

1. (1)

There is an isomorphism $\phi:A\to B$ such that $\phi(A\cap P)=B\cap P$; 2. (2)

Every left coset $gA\in G/A$ such that $gA\cap P\neq\emptyset$ has a minimal coset representative $p\in P$: $q\in gA\cap P\Rightarrow p\leq q$; 3. (3)

For every $x,y\in B$, $x\vee y<\infty\Rightarrow x\vee y\in B$.

Let $G^{*}=\langle G,t\mid t^{-1}at=\phi(a),a\in A\rangle$ be the HNN extension of $G$ and let $P^{*}$ be the subsemigroup of $G^{*}$ generated by $\{P,t\}$. Then $(G^{*},P^{*})$ is quasi-lattice ordered.

Before we can prove Theorem 4.1, we need to prove two lemmas. The first shows that elements of $P^{*}$ are guaranteed to have normal forms made up of elements of $P$ and $t$ if and only if condition (2) of Theorem 4.1 holds. The second is a technical lemma which we will use several times in Theorem 4.1 and in later proofs.

Lemma 4.2**.**

Suppose that $(G,P)$ is a quasi-lattice ordered group with subgroups $A$ and $B$. Suppose that $\phi:A\to B$ is a group isomorphism such that $\phi(A\cap P)=B\cap P$. Let $G^{*}=\langle G,t\mid t^{-1}at=\phi(a),a\in A\rangle$ be the corresponding HNN extension of $G$ and let $P^{*}$ be the subsemigroup of $G^{*}$ generated by $P\cup\{t\}$. Let

[TABLE]

The following two statements are equivalent:

1. (1)

Every left coset $gA\in G/A$ such that $gA\cap P\neq\emptyset$ has a minimal coset representative $p\in P$; 2. (2)

There exists a complete set $X$ of left coset representatives such that $L_{A}\subseteq X$ and every $\alpha\in P^{*}$ has normal form

[TABLE]

Proof.

Assume (1). Choose a complete set $X$ of coset representatives which contains $L_{A}$. Let $\alpha\in P^{*}$. If $\theta(\alpha)=0$ then $\alpha\in P$, and $\alpha$ has form (4.1) trivially.

We proceed by induction on $\theta(\alpha)\geq 1$. Suppose that $\theta(\alpha)=1$. We may write $\alpha=q_{0}tq_{1}$ for some $q_{0},q_{1}\in P$. Then $q_{0}A\cap P\neq\emptyset$, and there exists $p_{0}\in L_{A}$ such that $p_{0}A=q_{0}A$ and $p_{0}\leq q_{0}$. Thus$p_{0}^{-1}q_{0}\in P\cap A$. Hence $q_{0}=p_{0}a$ for some $a\in A\cap P$. Thus $\alpha$ has normal form

[TABLE]

Since $\phi(A\cap P)=B\cap P$ we have $\phi(a)\in P$ and so $X$ satisfies (2).

Suppose that all $\alpha$ with $1\leq\theta(\alpha)\leq k$ have normal form (4.1). Consider $\alpha$ with $\theta(\alpha)=k+1$. We write

[TABLE]

By assumption, we can write the first $2k+2$ terms of $\alpha$ in normal form

[TABLE]

where $p_{i}\in L_{A}$ for $0\leq i<k$ and $r_{k}\in P$. There exists $p_{k}\in L_{A}$ such that $r_{k}A=p_{k}A$ and $p_{k}\leq r_{k}$. As above, we can write $r_{k}=p_{k}a$ for some $a\in A\cap P$. Then

[TABLE]

We set $p_{k+1}=\phi(a)q_{k+1}$, which is in $P$ because $\phi(a)$ is. Then $\alpha=p_{0}tp_{1}t\ldots tp_{k}tp_{k+1}$ has form (4.1). By induction, every $\alpha$ has normal form (4.1). This implies (2).

For (2) $\Rightarrow$ (1) we argue by contradiction: we will assume (2) holds but (1) doesn’t. Let $X$ be a set of coset representatives satisfying (2), and suppose that there exists some coset $gA$ such that $gA\cap P\neq\emptyset$ which has no minimal coset representative. Let $p\in X$ be the coset representative of $gA$. Since $p$ is not minimal, there exists $q\in gA\cap P$ with $p\not\leq q$. Thus $p^{-1}q\not\in P$. Consider $qt\in P^{*}$ in normal form:

[TABLE]

Since $\phi(A\cap P)=B\cap P$ we have $\phi(p^{-1}q)\not\in P$. So (4.2) is not the normal form (4.1), a contradiction. Thus $(2)\Rightarrow(1)$. ∎

Lemma 4.3**.**

Let $(G,P)$ be a quasi-lattice ordered group and let $B$ be a subgroup of $G$. Suppose that for every $x,y\in B$, $x\vee y<\infty\Rightarrow x\vee y\in B$. Then for all $x\in B\cap PP^{-1}$, there exist $\mu,\nu\in B\cap P$ such that $x=\mu\nu^{-1}$ and for all $p,q\in P$ with $pq^{-1}=x$ we have $\mu\leq p$ and $\nu\leq q$.

The lemma says that if $x\in PP^{-1}\cap B$, then the minimal elements of (2.1) must also be contained in $B$. In particular, if $\phi(A\cap P)=B\cap P$, then $\phi(x)\in PP^{-1}$.

Proof of Lemma 4.3.

Fix $x\in B\cap PP^{-1}$. Say $x=st^{-1}$ with $s,t\in P$. Then $x^{-1}s\in P$ and $x\leq s$. Also $e\leq s$, and so $x\vee e<\infty$. Since $e,x\in B$ we get $x\vee e\in B$. Let $\mu=x\vee e$ and $\nu=x^{-1}(x\vee e)$. Then $\mu\nu^{-1}=x\vee e(x^{-1}(x\vee e))^{-1}=x$.

Fix $p,q\in P$ such that $x=pq^{-1}$. Then $x^{-1}p=(pq^{-1})^{-1}p=q$, and so $x\leq p$. Therefore $\mu=x\vee e\leq p$. Now $\mu^{-1}p\in P$, and then $\nu^{-1}q=\mu^{-1}p\in P$ gives $\nu\leq q$. ∎

We can now prove Theorem 4.1. Its proof is based on [11, Theorem 2.11], and our presentation is helped by Emily Irwin’s treatment of [11, Theorem 2.11] in her University of Otago Honours thesis.

Proof of Theorem 4.1.

Fix $x\in P^{*}P^{*-1}$. We shall prove that there exist $\mu,\nu\in P^{*}$ with $x=\mu\nu^{-1}$ such that whenever $\gamma\delta^{-1}=x$ we have $\mu\leq\gamma$ and $\nu\leq\delta$ (see (2.1)).

Choose $\alpha,\beta\in P^{*}$ such that $x=\alpha\beta^{-1}$. Choose a complete set $X$ of left coset representatives of $G/A$ that contains

[TABLE]

By Lemma 4.2 we can write $\alpha$ and $\beta$ in unique normal form:

[TABLE]

[TABLE]

Now $x=\alpha\beta^{-1}$ is equal to

[TABLE]

First we look for initial cancellations in the middle of $\alpha\beta^{-1}$: if $rs^{-1}\in B$, then we can replace $trs^{-1}t^{-1}$ with $\phi^{-1}(rs^{-1})$. By assumption (3), Lemma 4.3 applies and there exist $b_{1},b_{2}\in P\cap B$ such that $rs^{-1}=b_{1}b_{2}^{-1}$. Then

[TABLE]

Since $\phi(A\cap P)=B\cap P$ we have $\phi^{-1}(b_{1})\phi^{-1}(b_{2})^{-1}\in PP^{-1}$. Then

[TABLE]

We can repeat this process until there are no more cancellations available in the middle, and so we assume this is the case for the expression (4.3). This gives the following cases:

1. (1)

there are no $t$ and no more $t^{-1}$, 2. (2)

there are no more $t^{-1}$, 3. (3)

there are no more $t$, 4. (4)

there are $t$ and $t^{-1}$, and then the term with $t$ to the left and $t^{-1}$ to its right is not in $B$.

In each case, we will write down our candidates for $\mu$ and $\nu$ and prove that they are the required minimums.

(1) Suppose that after initial cancellations, there are no more $t$ and no more $t^{-1}$. Then $\alpha\beta^{-1}=p_{0}q_{0}^{-1}$ is already in normal form. By (2.1) there exist $\sigma,\tau\in P$ such that $p_{0}q_{0}^{-1}=\sigma\tau^{-1}$ and for all $c,d\in P$ such that $cd^{-1}=\sigma\tau^{-1}$ we have $\sigma\leq c$ and $\tau\leq d$. So we write $x=\sigma\tau^{-1}$ and choose as our candidates $\mu=\sigma$ and $\nu=\tau$.

Let $\gamma,\delta\in P^{*}$ such that $x=\gamma\delta^{-1}$. Let $\theta$ be the height map. Then $\theta(x)=0$ and hence $\theta(\gamma)=\theta(\delta)$. We will prove that $\mu\leq\gamma$ and $\nu\leq\delta$ by induction on $\theta(\gamma)$.

For $\theta(\gamma)=0$ we have $\gamma,\delta\in P$ , and then $\mu=\sigma\leq\gamma$ and $\nu=\tau\leq\delta$. Let $k\geq 0$ and suppose that for all $\gamma,\delta\in P^{*}$ such that $\theta(\gamma)=\theta(\delta)=k$ and $x=\gamma\delta^{-1}$ we have $\mu\leq\gamma$ and $\nu\leq\delta$. Now consider $\gamma,\delta\in P^{*}$ such that $x=\gamma\delta^{-1}$ and $\theta(\gamma)=\theta(\delta)=k+1$. We write $\gamma=m_{0}t\ldots m_{k}tm_{k+1}$ and $\delta=n_{0}t\ldots n_{k}tn_{k+1}$ in normal form where $m_{i},n_{i}\in L_{A}$ for $0\leq i\leq k$ and $m_{k+1},n_{k+1}\in P$. Next we reduce $x=\gamma\delta^{-1}$ towards normal form. We have

[TABLE]

Since $x$ has a unique normal form with no $t$ or $t^{-1}$, there must be some cancellation. Since the $m_{i},n_{i}\in L_{A}$ for $0\leq i\leq k$, the cancellation must occur across $m_{k+1}n_{k+1}^{-1}$. So $m_{k+1}n_{k+1}^{-1}\in B$ and $tm_{k+1}n_{k+1}^{-1}t^{-1}=\phi^{-1}(m_{k+1}n_{k+1}^{-1})$, and

[TABLE]

By assumption (3), Lemma 4.3 applies and there exists $b_{m},b_{n}\in B\cap P$ such that $m_{k+1}n_{k+1}^{-1}=b_{m}b_{n}^{-1}$ and $b_{m}\leq m_{k+1}$ and $b_{n}\leq n_{k+1}$. Then

[TABLE]

Since $\phi(A\cap P)=B\cap P$ we have $m_{k+1}\phi^{-1}(b_{m}),n_{k+1}\phi^{-1}(b_{n})\in P$. But now we have $\gamma^{\prime}=m_{0}t\ldots tm_{k}\phi^{-1}(b_{m})$ and $\delta^{\prime}=n_{0}t\ldots tn_{k}\phi^{-1}(b_{n})$ such that $\gamma^{\prime}(\delta^{\prime})^{-1}=x$ and $\theta(\gamma)=\theta(\delta^{\prime})=k$. By our induction hypothesis we have $\mu\leq\gamma^{\prime}$ and $\nu\leq\delta^{\prime}$.

To show that $\mu\leq\gamma$ we compute:

[TABLE]

Since $b_{m}^{-1}m_{k+1}\in P$ we have $tb_{m}^{-1}m_{k+1}\in P^{*}$. Now $(\gamma^{\prime})^{-1}\gamma\in P^{*}$ and $\gamma^{\prime}\leq\gamma$. Since $\mu\leq\gamma^{\prime}$ this gives $\mu\leq\gamma$. Similarly, $\delta=\delta^{\prime}tb_{n}^{-1}n_{k+1}$ and so $\nu\leq\delta$. By induction, for all $\gamma,\delta\in P^{*}$ such that $x=\gamma\delta^{-1}$ we have $\mu\leq\gamma$ and $\nu\leq\delta$.

(2) Suppose that after the initial cancellations there are no more $t^{-1}$ left. Then we have $x$ in normal form:

[TABLE]

where $m\geq n$. By (2.1) there exist $\sigma,\tau\in P$ such that $rs^{-1}=\sigma\tau^{-1}$ and $\sigma\leq r$ and $\tau\leq s$. So

[TABLE]

Our candidates are

[TABLE]

Fix $\gamma,\delta\in P^{*}$ such that $x=\gamma\delta^{-1}$. Say $\gamma=m_{0}tm_{1}\ldots tm_{i}$ and $\delta=n_{0}tn_{1}\ldots n_{j}$ in normal form. Then

[TABLE]

Since $\theta(\gamma\delta^{-1})=\theta(\mu\nu^{-1})=m-n$ we get

[TABLE]

and hence $i\geq m-n$. It follows from the uniqueness of normal form that there exists $\gamma^{\prime}\in P^{*}$ such that

[TABLE]

Thus $p_{0}tp_{1}t\ldots tp_{m-n-1}t\leq\gamma$. From (4.4) we have $(p_{0}tp_{1}t\ldots tp_{m-n-1}t)^{-1}x=\sigma\tau^{-1}$. Since $(p_{0}tp_{1}t\ldots tp_{m-n-1}t)^{-1}\gamma,\delta\in P^{*}$ and $(p_{0}tp_{1}t\ldots tp_{m-n-1}t)^{-1}\gamma\delta=\sigma\tau^{-1}$, we can apply (1) above with $\mu^{\prime}=\sigma$ and $\nu^{\prime}=\tau$ to see that $\sigma\leq(p_{0}tp_{1}t\ldots tp_{m-n-1}t)^{-1}\gamma$ and $\tau\leq\delta$. Hence $\mu=p_{0}tp_{1}t\ldots tp_{m-n-1}t\sigma\leq\gamma$ and $\nu\leq\delta$ as required.

(3) Suppose that after the cancellations, there are no more $t$ left. Then

[TABLE]

for some $r,s\in P$. Consider

[TABLE]

By (2.1) there exist $\sigma,\tau\in P$ such that $rs^{-1}=\sigma\tau^{-1}$ and $\sigma\leq r$ and $\tau\leq s$. Let

[TABLE]

By (2) they have the property that $x^{-1}=\nu\mu^{-1}$ and for all $\gamma,\delta\in P^{*}$ such that $x^{-1}=\delta\gamma^{-1}$ we have $\mu\leq\gamma$ and $\nu\leq\delta$. Taking inverses, $x=\mu\nu^{-1}$ and for all $\gamma,\delta\in P^{*}$ such that $x=\gamma\delta^{-1}$ we have $\mu\leq\gamma$ and $\nu\leq\delta$.

(4) Suppose that after the initial cancellations there are both $t$ and $t^{-1}$ left. Then the term with $t$ to the left and $t^{-1}$ to its right is not in $B$. There exist $k\leq m$ and $l\leq n$ such that

[TABLE]

By (2.1) there exist $\sigma,\tau\in P$ such that $rs^{-1}=\sigma\tau^{-1}$ and $\sigma\leq r$ and $\tau\leq s$. Our candidate for $\mu,\nu$ are

[TABLE]

Fix $\gamma,\delta\in P^{*}$ such that $x=\gamma\delta^{-1}$. By the argument used in (2), there exists $\gamma^{\prime}\in P^{*}$ such that $\gamma\delta^{-1}=p_{0}tp_{1}t\ldots tp_{j}t\gamma^{\prime}\delta^{-1}$. Hence $\gamma^{\prime}=(p_{0}tp_{1}t\ldots tp_{j}t)^{-1}\gamma\in P^{*}$ and $p_{0}tp_{1}t\ldots tp_{j}t\leq\gamma$.

Consider $\gamma^{\prime}\delta^{-1}=\sigma\tau^{-1}t^{-1}q_{k}^{-1}t^{-1}\ldots t^{-1}q_{0}^{-1}$. Here $\gamma^{\prime},\delta\in P^{*}$ and there are no $t$ in $\gamma^{\prime}\delta^{-1}$ after cancellation. Applying (3) with $\mu^{\prime}=\sigma$ and $\nu^{\prime}=\nu$ to get $\mu^{\prime}=\sigma\leq\gamma^{\prime}$ and $\nu^{\prime}=\nu\leq\delta$. Then $\mu=p_{0}tp_{1}t\ldots tp_{j}t\sigma\leq p_{0}tp_{1}t\ldots tp_{j}t\gamma^{\prime}=\gamma$. ∎

Theorem 4.1 gives new examples of quasi-lattice ordered HNN extensions.

Example 4.4**.**

We can show that the Baumslag-Solitar group $(\operatorname{BS}(c,d),\operatorname{BS}(c,d)^{+})$ with $c,d>0$ is quasi-lattice ordered using Theorem 4.1. Since $(\mathbb{Z},\mathbb{N})$ is totally ordered it is quasi-lattice ordered. Let $A=\{dm:m\in\mathbb{Z}\}$ and $B=\{cm:m\in\mathbb{Z}\}$. Every element $n\in\mathbb{N}$ has a unique decomposition $n=r+md$ where $m\in\mathbb{N}$ and $r\in\{0,1,\ldots,d-1\}$. The remainder $r$ is a choice of coset representative $n+A=r+A$. For all $n^{\prime}\in n+A\cap\mathbb{N}$ we have $n^{\prime}=r+md+kd$ where $k\in\mathbb{Z}$ and $k\leq m$. Thus $r\leq n^{\prime}$. Hence every coset of $\mathbb{Z}/A$ has nontrivial intersection with $\mathbb{N}$, and has a minimal coset representative in $\mathbb{N}$. Since $B$ is totally ordered it is closed under taking least upper bounds. Define $\phi:A\to B$ by $\phi(dm)=cm$. Then $\phi(A\cap\mathbb{N})=B\cap\mathbb{N}$. So Theorem 4.1 applies and gives that $(\mathbb{Z}^{*},\mathbb{N}^{*})$ is quasi-lattice ordered.

Example 4.5**.**

We can generalise the previous example to $(\mathbb{Z}^{2},\mathbb{N}^{2})$, which is quasi-lattice ordered by [9, Example 2.3(2)]. Fix $a,b,c,d\in\mathbb{N}$. Then $A=\{(am,bn):m,n\in\mathbb{Z}\}$ and $B=\{(cm,dn):m,n\in\mathbb{Z}\}$ are subgroups of $\mathbb{N}^{2}$. Let $\phi:A\to B$ be defined by $\phi((am,bn))=(cm,dn)$. This $\phi$ satisfies $\phi(A\cap\mathbb{N}^{2})=B\cap\mathbb{N}^{2}$. For all $(m,n)\in\mathbb{N}^{2}$, the division algorithm on $\mathbb{N}$ gives a unique decomposition

[TABLE]

Thus $(r_{1},r_{2})$ is a minimal left coset representative of $(m,n)+A$. For all $(m,n),(p,q)\in B$, we have

[TABLE]

and hence $(m,n)\vee(p,q)\in B$. So $B$ is closed under $\vee$. By Theorem 4.1, $(\mathbb{Z}^{2*},\mathbb{N}^{2*})$ is a quasi-lattice ordered group with presentation

[TABLE]

It is straightforward to extend this construction to $(\mathbb{Z}^{n},\mathbb{N}^{n})$.

Example 4.6**.**

Consider the free group $\mathbb{F}_{2}$ on $2$ generators $\{a,b\}$ and let $\mathbb{F}_{2}^{+}$ be the subsemigroup generated by $e$, $a$ and $b$. The pair $(\mathbb{F}_{2},\mathbb{F}_{2}^{+})$ is quasi-lattice ordered by [9, Example 2.3(4)]. Let $A=\{a^{n}:n\in\mathbb{Z}\}$, $B=\{b^{n}:n\in\mathbb{Z}\}$ and $\phi:A\to B$ defined by $\phi(a^{n})=b^{n}$. Every $x\in F_{2}^{+}$ can be written as a product of $y\in\mathbb{F}_{2}^{+}$ which does not end in $a$ followed by $a^{n}$ for some $n\geq 0$. Then $y\in xA$. Every $z\in yA\cap\mathbb{F}_{2}^{+}$ begins with the word $y$ which is in $\mathbb{F}_{2}^{+}$. It follows that $y\leq z$. Thus $A$ has minimal left coset representatives in $\mathbb{F}_{2}^{+}$. Since $B$ is totally ordered, it is trivially closed under $\vee$. It follows from Theorem 4.1 that $(\mathbb{F}_{2}^{*},\mathbb{F}_{2}^{+*})$ is a quasi-lattice ordered group with presentation

[TABLE]

Example 4.7**.**

Building on $(\mathbb{F}_{2},\mathbb{F}_{2}^{+})$ again, fix $s,u\in\mathbb{N}$, and let $A=\{a^{ms}:m\in\mathbb{Z}\}$, $B=\{b^{mu}:m\in\mathbb{Z}\}$ and $\phi:A\to B$ be $\phi(a^{ms})=b^{mu}$. Then $B$ is totally ordered and hence is closed under $\vee$. To see that $A$ has minimal coset representatives, we observe that every $x\in\mathbb{F}_{2}^{+}$ is a product of a $y\in\mathbb{F}_{2}^{+}$ that does not end in $a$ followed by $a^{n}$ for some $n\in\mathbb{N}$. We write $n=r+js$ for some $j\in\mathbb{N}$ and $r\in\{0,1,\ldots,s-1\}$. We choose $ya^{r}$ as our coset representative. Then for all $z\in ya^{r}A\cap\mathbb{F}_{2}^{+}$ we have $ya^{r}\leq z$. It follows from Theorem 4.1 that $(\mathbb{F}_{2}^{*},\mathbb{F}_{2}^{+*})$ is a quasi-lattice ordered group with presentation

[TABLE]

Taking $u=s=1$ gives Example 4.6.

Replacing $B$ by $B^{\prime}=\{a^{mu}:m\in\mathbb{Z}\}$ gives a quasi-lattice ordered group $(\mathbb{F}_{2}^{*},\mathbb{F}_{2}^{+*})$ with presentation

[TABLE]

In the next two examples we show that it is easy to find subgroups which do not have minimal left coset representatives.

Example 4.8**.**

Consider the group

[TABLE]

with subsemigroup $\mathbb{Z}(\sqrt{2})^{+}=\mathbb{Z}(\sqrt{2})\cap[0,\infty)$. Let $A=\mathbb{Z}(2\sqrt{2})$. We claim there are no minimal coset representatives for $A$. Suppose, aiming for a contradiction, that there exists some coset representative $p\in\mathbb{Z}(\sqrt{2})^{+}$ such that

[TABLE]

Recall that $\mathbb{Z}(2\sqrt{2})$ is dense in $\mathbb{R}$.222To see denseness observe that $0<(-2+2\sqrt{2})<1$ and $(-2+2\sqrt{2})^{n}\in\mathbb{Z}(\sqrt{2})$ for all $n\in\mathbb{N}$. Thus for every open interval $(u,v)$ there exists $n$ such that $(-2+2\sqrt{2})^{n}<v-u$. Hence there exists $k\in\mathbb{N}$ such that $k(-2+2\sqrt{2})^{n}\in(u,v)$. Thus there exists some $a\in\mathbb{Z}(2\sqrt{2})\cap(0,p)$ hence $a\leq p$. Thus $p-a\in[p+\mathbb{Z}(2\sqrt{2})]\cap\mathbb{Z}(\sqrt{2})^{+}$. But $p-a\leq p$, giving a contradiction.

Example 4.9**.**

Consider $(\mathbb{Z}^{2},\mathbb{N}^{2})$, and let $A$ be the subgroup generated by $\{(1,2),(2,1)\}$. Consider the coset

[TABLE]

Since $(1,0)$ and $(0,1)$ have no nonzero lower bound, there can be no choice of minimal coset representative.

5. Amenability of $(G^{*},P^{*})$

In this section we prove the following theorem.

Theorem 5.1**.**

Let $(G,P)$ be a quasi-lattice ordered group with subgroups $A$ and $B$. Suppose that $\phi:A\to B$ is an isomorphism which satisfies the hypotheses of Theorem 4.1. Let $(G^{*},P^{*})$ be the corresponding HNN extension. If $(G,P)$ is amenable, then $(G^{*},P^{*})$ is amenable.

To prove the theorem we will show that the height map $\theta:G^{*}\to\mathbb{Z}$ is a controlled map, that $(\theta^{-1}(e),\theta^{-1}(e)\cap P^{*})$ is amenable, and then apply Theorem 3.2. To prove that $(\theta^{-1}(e),\theta^{-1}(e)\cap P^{*})$ is amenable, we start by investigating order-preserving isomorphisms between the semigroups of quasi-lattice ordered groups.

Lemma 5.2**.**

Let $(G,P)$ and $(K,Q)$ be quasi-lattice ordered groups. Suppose that there is a semigroup isomorphism $\phi:P\to Q$. Then $\phi$ is order-preserving. In particular, for $x,y\in P$, $x\vee y<\infty$ if and only if $\phi(x)\vee\phi(y)<\infty$. If $x\vee y<\infty$ then $\phi(x\vee y)=\phi(x)\vee\phi(y)$.

Proof.

To see that $\phi$ is order-preserving, let $x,y\in P$ and $x\leq y$. Then $y\in xP$, and $\phi(y)\in\phi(x)\phi(P)=\phi(x)Q$. Thus $\phi(x)\leq\phi(y)$, and $\phi$ is order-preserving.

To show that $\phi$ preserves the least upper bound structure, first suppose that $x,y\in P$ such that $x\vee y<\infty$. Since $\phi$ is order-preserving it follows that $\phi(x),\phi(y)\leq\phi(x\vee y)$. Thus $\phi(x),\phi(y)$ have a common upper bound in $Q$. Hence $\phi(x)\vee\phi(y)$ exists and $\phi(x)\vee\phi(y)\leq\phi(x\vee y)$. Second, suppose that $\phi(x)\vee\phi(y)<\infty$ for some $x,y\in P$. Since $\phi^{-1}:Q\to P$ is a semigroup isomorphism it is order-preserving. Thus $\phi^{-1}(\phi(x)\vee\phi(y))$ is an upper bound for $x=\phi^{-1}(\phi(x))$ and $y=\phi^{-1}(\phi(y))$. Hence $x\vee y$ exists and $x\vee y\leq\phi^{-1}(\phi(x)\vee\phi(y))$. Thus

[TABLE]

Hence $x\vee y<\infty$ if and only if $\phi(x)\vee\phi(y)<\infty$, and $\phi(x\vee y)=\phi(x)\vee\phi(y)$. ∎

Proposition 5.3**.**

Let $(G,P)$ and $(K,Q)$ be quasi-lattice ordered groups. Let $\{v_{p}:p\in P\}$ and $\{w_{q}:q\in Q\}$ be the generating elements of $C^{*}(G,P)$ and $C^{*}(K,Q)$, respectively. Suppose that there is a semigroup isomorphism $\phi:P\to Q$.

1. (1)

There exists an isomorphism $\pi_{\phi}:C^{*}(G,P)\to C^{*}(K,Q)$ such that $\pi_{\phi}(v_{p})=w_{\phi(p)}$. 2. (2)

$(G,P)$* is amenable if and only if $(K,Q)$ is amenable.*

Proof.

(1) We will show that $T:P\to C^{*}(K,Q)$ defined by $T_{p}=w_{\phi(p)}$ is a covariant representation of $P$, and then take $\pi_{\phi}:=\pi_{T}$. Fix $p,q\in P$. Since $\phi$ is a semigroup isomorphism we have

[TABLE]

and $T_{e}=w_{\phi(e_{G})}=w_{e_{K}}=1$. Hence $T$ is an isometric representation. We have

[TABLE]

Hence $T$ is covariant. By the universal property of $C^{*}(G,P)$ there exists a homomorphism $\pi_{\phi}:C^{*}(G,P)\to C^{*}(K,Q)$ such that $\pi_{\phi}(v_{p})=T_{p}=w_{\phi(p)}$.

Since $\phi^{-1}:Q\to P$ is an isomorphism the argument above gives a homomorphism $\pi_{\phi^{-1}}:C^{*}(K,Q)\to C^{*}(G,P)$ such that $\pi_{\phi^{-1}}(w_{q})=v_{\phi^{-1}(q)}$. In particular,

[TABLE]

and $\pi_{\phi}(\pi_{\phi^{-1}}(w_{q}))=w_{q}$ It follows that $\pi_{\phi}$ is an isomorphism from $C^{*}(G,P)$ to $C^{*}(K,Q)$.

(2) By symmetry it suffices to show that if $(K,Q)$ is amenable then $(G,P)$ is amenable. Let $E_{Q}$ and $E_{P}$ be the conditional expectations on $C^{*}(K,Q)$ and $C^{*}(G,P)$, respectively. To see $E_{P}=\pi_{\phi}^{-1}\circ E_{Q}\circ\pi_{\phi}$ we compute on spanning elements:

[TABLE]

Suppose that $(K,Q)$ is amenable. Then $E_{Q}$ is faithful. Suppose that $a\geq 0$ and $E_{P}(a)=0$. Then $\pi_{\phi}(a)\geq 0$, and

[TABLE]

because $E_{Q}$ is faithful. Since $\pi_{\phi}$ is faithful, $a=0$. Now $E_{P}$ is faithful, and hence $(G,P)$ is amenable. ∎

Next we need some lemmas which will be used to show that the height map $\theta$ is a controlled map. In particular we need to identify the minimal elements of Definition 3.1. If $x\in P^{*}$ has normal form

[TABLE]

we call $p_{0}tp_{1}t\ldots p_{n-1}t$ the stem of $x$ and write

[TABLE]

The set of stems is our candidate for the minimal elements.

Lemma 5.4**.**

Let $(G,P)$ be a quasi-lattice ordered group with subgroups $A$ and $B$. Suppose that $\phi:A\to B$ is an isomorphism which satisfies the hypotheses of Theorem 4.1. Let $p,q\in P$. Then $p$ and $q$ have a common upper bound in $P^{*}$ if and only if $p$ and $q$ have a common upper bound in $P$.

Proof.

First suppose that $p$ and $q$ have a common upper bound $r\in P$. Then $r\in P^{*}$ and so $r$ is a common upper bound for $p$ and $q$ in $P^{*}$.

Second, suppose that $p$ and $q$ have a common upper bound $x\in P^{*}$. If $\theta(x)=0$, then $x\in P$ and we are done. Suppose, aiming for a contradiction, that $\theta(x)=k$ for some $k\geq 1$, and that $p,q$ have no common upper bound $y$ with $\theta(y)<k$.

Observe that $p^{-1}x,q^{-1}x\in P^{*}$, and that $\theta(p^{-1}x)=\theta(x)=\theta(q^{-1}x)=k$. We write $p^{-1}x$ and $q^{-1}x$ in their normal forms:

[TABLE]

where $p_{i},q_{i}\in L_{A}$ for $i<k$ and $p_{k},q_{k}\in P$. Consider

[TABLE]

Since $p^{-1}x(q^{-1}x)^{-1}=p^{-1}q$ and $p^{-1}q$ is in normal form we must have some cancellation. Since the first $k$ terms are already in normal form, $p_{k}q_{k}^{-1}\in B$. By Lemma 4.3, there exist $b_{1},b_{2}\in B\cap P$ such that $b_{1}\leq p_{k}$, $b_{2}\leq q_{k}$ and $b_{1}b_{2}^{-1}=p_{k}q_{k}^{-1}$. Then

[TABLE]

Rearranging gives

[TABLE]

Therefore $y=p(p_{0}tp_{1}t\ldots p_{k-1}\phi^{-1}(b_{1}))$ is a common upper bound for $p$ and $q$ in $P^{*}$ and $\theta(y)=k-1$, giving us the contradiction we sought. Therefore $p$ and $q$ have a common upper bound $y$ with $\theta(y)=0$, and hence they have a common upper bound in $P$. ∎

The statement of Lemma 5.5 is adapted from [2, Lemma 3.4].

Lemma 5.5**.**

Let $x,y\in P^{*}$ such that $x\vee y<\infty$. Write

[TABLE]

1. (1)

If $\theta(x)=\theta(y)$, then $\operatorname{stem}(x)=\operatorname{stem}(y)$ and $p\vee q<\infty$. In particular, $x\vee y=\operatorname{stem}(x)(p\vee q)$ and $\theta(x\vee y)=\theta(x)=\theta(y)$. 2. (2)

If $\theta(x)<\theta(y)$, then there exists $r\in P$ such that $x\vee y=yr$ and $\theta(x\vee y)=\theta(y)$.

In particular, $\theta(x\vee y)=\max\{\theta(x),\theta(y)\}$.

Proof.

(1) Suppose that $\theta(x)=\theta(y)$. We know that $x\leq x\vee y$ and $y\leq x\vee y$. Thus, by the uniqueness of normal forms, $\operatorname{stem}(x)=\operatorname{stem}(y)$. Now by left invariance of the partial order we see that

[TABLE]

Therefore $p$ and $q$ have a common left upper bound in $P^{*}$ and hence, by Lemma 5.4, they have a common left upper bound in $P$ and $p\vee q$ exists in $P$. By left invariance $x\vee y=\operatorname{stem}(x)p\vee q$. Further, $\theta(x\vee y)=\theta(x)=\theta(y)$.

(2) Suppose that $\theta(x)<\theta(y)$. Since $x\leq x\vee y$ we have $x^{-1}(x\vee y)\in P^{*}$. We can write $x^{-1}(x\vee y)=\tau\gamma u$ for some $u\in P$, $\tau\in\Sigma_{\theta(y)-\theta(x)}$ and $\gamma\in\Sigma_{\theta(x\vee y)-\theta(y)}$. Then $x\vee y=x\tau\gamma u$.

Now we have $x\tau\leq x\vee y$ and $\theta(x\tau)=\theta(x)+(\theta(y)-\theta(x))=\theta(y)$. Write $x\tau=\operatorname{stem}(x\tau)w$ for some $w\in P$. Therefore $x\tau\vee y<\infty$ and $\theta(x\tau)=\theta(y)$ so we can apply (1) to see that $\operatorname{stem}(x\tau)=\operatorname{stem}(y)$ and $x\tau\vee y=\operatorname{stem}(y)(q\vee w)$.

Now $x\vee y\leq\operatorname{stem}(y)(q\vee w)$. Therefore there exists some $r\in P$ such that $x\vee y=\operatorname{stem}(y)qr=yr$. Then $\theta(x\vee y)=\theta(y)$.

By (1) and (2) we see that

[TABLE]

Thus $\theta(x\vee y)=\max\{\theta(x),\theta(y)\}$. ∎

Proof of Theorem 5.1.

We will use Theorem 3.2; to do so we need to show that the height map $\theta:(G^{*},P^{*})\to(\mathbb{Z},\mathbb{N})$ is a controlled map in the sense of Definition 3.1, and that $(\theta^{-1}(e),\theta^{-1}(e)\cap P^{*})$ is amenable.

To see that $\theta$ is order-preserving, let $x,y\in P^{*}$ such that $x\leq y$. Then $x^{-1}y\in P^{*}$ and $\theta(x^{-1}y)\geq 0$. So $0\leq\theta(x^{-1}y)=-\theta(x)+\theta(y)$ and hence $\theta(x)\leq\theta(y)$. By Lemma 5.5, if $x\vee y<\infty$, then $\theta(x\vee y)=\max\{\theta(x),\theta(y)\}=\theta(x)\vee\theta(y)$.

For every $k\in\mathbb{N}$, $\Sigma_{k}$ is complete: if $x\in\theta^{-1}(k)\cap P^{*}$, then $\operatorname{stem}(x)\in\Sigma_{k}$ and $x=\operatorname{stem}(x)p$ for some $p\in P$. Hence $\operatorname{stem}(x)\leq x$. By the uniqueness of normal forms, if $\sigma,\tau\in\Sigma_{k}$ and $\sigma\vee\tau<\infty$ then $\sigma=\tau$. Therefore $\theta$ is a controlled map into the amenable group $\mathbb{Z}$.

Suppose that $(G,P)$ is amenable. Then $\theta^{-1}(0)\cap P^{*}$ is the set of elements of $P^{*}$ with height [math], and hence they all have normal form $p_{0}$ for some $p_{0}\in P$. Thus $\theta^{-1}(0)\cap P^{*}$ is isomorphic to $P$. Since $(G,P)$ is amenable, so is $(\theta^{-1}(0),\theta^{-1}(0)\cap P^{*})$ by Lemma 5.3. Since $(\mathbb{Z},\mathbb{N})$ is amenable, it now follows from Theorem 3.2 that $(G^{*},P^{*})$ is an amenable quasi-lattice ordered group. ∎

Example 5.6**.**

Since $(\mathbb{Z}^{2},\mathbb{N}^{2})$ and $(\mathbb{F}_{2},\mathbb{F}_{2}^{+})$ are amenable quasi-lattice ordered groups [9, §5.1], Theorem 5.1 implies that the HNN extensions $(\mathbb{Z}^{2*},\mathbb{N}^{2*})$ and $(\mathbb{F}_{2}^{*},\mathbb{F}_{2}^{+*})$ in Examples 4.5-4.7 are amenable quasi-lattice ordered groups.