2. A groupoid model for $C_{c}^{*}(P)$

In this section, we review the groupoid model for $C_{c}^{*}(P)$ described in [Reference Salas24] and in [Reference Sundar25]. We follow the exposition given in [Reference Sundar25]. Let $G$ be a countable, discrete abelian group, and let $P$ be a subsemigroup of $G$ containing the identity element $0$ . We assume that $P$ generates $G$ , i.e. $P-P=G$ .

Recall that $C_{c}^{*}(P)$ is the universal unital $C^{*}$ -algebra generated by a family of isometries $\{v_a\,:\, a \in P\}$ such that

1. (1) for $a,b \in P$ , $v_av_b=v_{a+b}$ , and

2. (2) for $a,b \in P$ , $e_ae_b=e_be_a$ , where $e_a\,:\!=\,v_av_a^{*}$ and $e_b\,:\!=\,v_bv_b^{*}$ .

Let $s \in G$ be given. Choose $a,b\in P$ such that $s=a-b$ . Set $w_s\,:\!=\,v_b^{*}v_a$ . Thanks to Proposition 3.4 of [Reference Sundar25], $w_s$ is well defined, and $\{w_s\,:\, s \in G\}$ is a family of partial isometries whose range projections commute. For $s \in G$ , let $e_s\,:\!=\,w_sw_s^*$ .

Next, we describe a groupoid whose $C^{*}$ -algebra is a quotient of $C_{c}^{*}(P)$ . Let $\mathcal{P}(G)$ be the power set of $G$ which we identify, in the usual way, with $\{0,1\}^{G}$ . Endow $\mathcal{P}(G)$ with the product topology inherited via this identification. Then, $\mathcal{P}(G)$ is a compact Hausdorff space and is metrisable. The map

\begin{equation*}\mathcal{P}(G) \times G \ni (A,s) \to A+s \in \mathcal{P}(G)\end{equation*}

defines an action of $G$ on $\mathcal{P}(G)$ .

Let

\begin{equation*}\overline {X}_{u}\,:\!=\{A \in \mathcal {P}(G)\,:\, 0 \in A, -P+A \subset A\}.\end{equation*}

Note that $\overline{X}_u$ is a closed subset of $\mathcal{P}(G)$ and hence a compact subset of $\mathcal{P}(G)$ . Also, $\overline{X}_u$ is $P$ -invariant, i.e. if $A \in \overline{X}_u$ and $a \in P$ , then $A+a \in \overline{X}_u$ . Let $\mathcal{G}$ be the reduction of the transformation groupoid $\mathcal{P}(G) \rtimes G$ onto $\overline{X}_u$ , i.e.

\begin{equation*} \mathcal {G}\,:\!=\{(A,s) \in \mathcal {P}(G) \times G\,:\, A \in \overline {X}_u, A+s \in \overline {X}_u\}=\{(A,s)\,:\, A \in \overline {X}_u, -s \in A\}.\end{equation*}

The multiplication and inversion on $\mathcal{G}$ are given by

\begin{align*} (A,s)(B,t)&\,:\!=\,(A,s+t) \,\,\,\,\,\,\,\,\textrm{if}\ A+s=B, \,\,\textrm{and}\\ (A,s)^{-1}&\,:\!=\,(A+s,-s). \end{align*}

The groupoid $\mathcal{G}$ is étale and is the usual Deaconu-Renault groupoid $\overline{X}_u \rtimes P$ .

For $f \in C(\overline{X}_u)$ , define $\widetilde{f} \in C_{c}(\overline{X}_u \rtimes P)$ by

(2.1) \begin{equation} \widetilde{f}(A,s)\,:\!=\,\begin{cases} f(A) & \mbox{ if } s =0, \\ 0 & \mbox{ if $s \neq 0$. } \end{cases} \end{equation}

Observe that $C(\overline{X}_u) \ni f \to \widetilde{f} \in C_{c}(\overline{X}_u \rtimes P)$ is a unital $^*$ -algebra homomorphism which is injective. Via this embedding, we identify $C(\overline{X}_u)$ as a unital $^*$ -subalgebra of $C_{c}(\mathcal{G})$ . For $f \in C(\overline{X}_u)$ , we abuse notation and we denote $\widetilde{f}$ by $f$ .

For $s \in G$ , let $\overline{w}_s \in C_{c}(\overline{X}_u \rtimes P)$ be defined by

(2.2) \begin{equation} \overline{w}_s(A,t)\,:\!=\,\begin{cases} 1 & \mbox{ if } t=-s, \\ 0 & \mbox{ if $t \neq -s$}. \end{cases} \end{equation}

For $a \in P$ , set $\overline{v}_a\,:\!=\,\overline{w}_a$ . Observe the following facts. The details involving routine computations are left to the reader.

1. (1) For $a \in P$ , $\overline{v}_a$ is an isometry and $\overline{v}_a\overline{v}_b=\overline{v}_{a+b}$ .

2. (2) For $a,b \in P$ and $s=a-b$ , $\overline{w}_s=\overline{v}_b^{*}\overline{v}_a$ .

3. (3) For $s \in G$ , let $\epsilon _{s} \in C(\overline{X}_u)$ be defined by the equation $\epsilon _{s}(A)=1_{A}(s)$ . Then, $\overline{w}_s\overline{w}_s^{*}=\epsilon _{s}$ for every $s \in G$ . Consequently, $\{\overline{w}_s\overline{w}_s^{*}\,:\,s \in G\}$ generates $C(\overline{X}_u)$ . Also, the range projections $\{\overline{v}_a\overline{v}_a^{*}\,:\,a \in P\}$ form a commuting family.

4. (4) We claim that the $C^{*}$ -algebra generated by $\{\overline{v}_a\,:\,a \in P\}$ is $C^{*}(\mathcal{G})$ . Let $f \in C_{c}(\mathcal{G})=C_c(\overline{X}_u \rtimes P)$ . It suffices to consider the case when $f$ is supported on $\overline{X}_u \times \{-s\}$ for some $s \in G$ . Let $h\,:\,\overline{X}_u \to \mathbb{C}$ be defined by

   \begin{equation*} h(A)\,:\!=\,\begin {cases} f(A,-s) & \mbox { if } (A,-s) \in \overline {X}_u \rtimes P, \\ 0 & \mbox { otherwise}. \end {cases} \end{equation*}
   Then, $h$ is continuous and $h*\overline{w}_s=f$ . By $(3)$ , $h$ lies in the $C^{*}$ -algebra generated by $\{\overline{w}_s\,:\,s \in G\}$ . Therefore, $\{\overline{w}_s\,:\,s \in G\}$ generates $C^{*}(\mathcal{G})$ . It follows from $(2)$ that $\{\overline{v}_a\,:\, a \in P\}$ generates the $C^{*}$ -algebra $C^{*}(\overline{X}_u \rtimes P)$ .

Hence, there exists a unique surjective $^*$ -homomorphism $\Phi \,:\,C_{c}^{*}(P) \to C^{*}(\overline{X}_u \rtimes P)$ such that

(2.3) \begin{equation} \Phi (v_a)=\overline{v}_a. \end{equation}

Theorem 7.4 of [Reference Sundar25] asserts that $\Phi$ is invertible, and $\Phi$ is an isomorphism. Via the map $\Phi$ , we identify $C_{c}^{*}(P)$ with $C^{*}(\overline{X}_u \rtimes P)$ . Henceforth, we do not distinguish between $w_s$ and $\overline{w}_s$ .

For $f \in C(\overline{X}_u)$ and $s \in G$ , let $R_{s}(f) \in C(\overline{X}_u)$ be defined by the equation

\begin{equation*} R_s(f)(A)\,:\!=\,\begin {cases} f(A-s) & \mbox{if} \, A-s \in \overline {X}_u, \\ 0 & \mbox { otherwise}. \end {cases} \end{equation*}

For $s \in G$ and $f \in C(\overline{X}_u) \subset C^{*}(\overline{X}_u \rtimes P)$ , the covariance relation

\begin{equation*}w_sfw_s^\ast=R_s(f),\end{equation*}

is satisfied. A pleasant consequence of the above covariance relation is the fact that the linear span of $\{fw_s\,:\, f \in C(\overline{X}_u), s \in G\}$ is a unital dense $^*$ -subalgebra of $C^{*}(\overline{X}_u \rtimes P)$ .

Let

\begin{equation*}\overline {Y}_{u}\,:\!=\{A \in \mathcal {P}(G)\,:\, A \neq \emptyset, -P+A \subset A\}.\end{equation*}

Observe that $\overline{Y}_u$ is a locally compact Hausdorff space and $G$ leaves $\overline{Y}_u$ invariant. Also, $\overline{X}_u$ is a clopen set in $\overline{Y}_u$ . It is clear that the groupoid $\overline{X}_u \rtimes P$ is the reduction of the transformation groupoid $\overline{Y}_u \rtimes G$ onto $\overline{X}_u$ . This has the consequence that $C^{*}(\overline{X}_u \rtimes P)$ is isomorphic to the cut-down $p(C_0(\overline{Y}_u) \rtimes G)p$ where $p=1_{\overline{X}_u}$ . Moreover, the union $\displaystyle \bigcup _{a \in P}(\overline{X}_u-a)=\overline{Y}_u$ . Consequently, it follows that $p$ is a full projection in $C_0(\overline{Y}_u) \rtimes G$ .

Often, we embed $C_{c}^{*}(P)$ inside the crossed product $C_0(\overline{Y}_u) \rtimes G$ and view $C_{c}^{*}(P)$ as a full corner of $C_{0}(\overline{Y}_u) \rtimes G$ . The embedding $C_{c}^{*}(P) \to C_0(\overline{Y}_u) \rtimes G$ is given by the rules

\begin{equation*}C(\overline {X}_u) \ni f \to f1_{\overline {X}_u} \in C_0(\overline {Y}_u) \textrm{and} \, w_s \to u_s1_{\overline {X}_u}.\end{equation*}

Here, $\{u_s\,:\, s \in G\}$ are the canonical unitaries of the crossed product $C_0(\overline{Y}_u)\rtimes G$ .

Tracial states on $C_{c}^{*}(P)$ : We first determine the tracial states on $C_c^{*}(P)$ . Let $C^*(G)$ be the group $C^{*}$ -algebra of $G$ , and let $\{u_s\,:\,s \in G\}$ be the canonical unitaries of $C^{*}(G)$ . By the universal property, there exists a surjective $^*$ -homomorphism $\pi \,:\,C_c^{*}(P) \to C^{*}(G)$ such that $\pi (v_a)=u_a$ . In the groupoid picture, $\pi$ coincides with the restriction map $Res\,:\,C^{*}(\overline{X}_u \rtimes P) \to C^{*}(\overline{X}_u \rtimes P|_{\{G\}}) \cong C^{*}(G)$ . Note that $\{G\}$ is an invariant closed subset of the unit space $\overline{X}_u$ of the groupoid $\mathcal{G}=\overline{X}_u \rtimes P$ .

Proof. Let $F\,:\!=\{G\}$ and $X_u\,:\!=\,\overline{X}_u\backslash F$ . Let $\omega$ be a tracial state on $C_{c}^{*}(P)$ . Let $m$ be the measure on $\overline{X}_u$ that corresponds to the state $\omega |_{C(\overline{X}_u)}$ . Let $a \in P$ be given. Note that $v_av_a^{*}=1_{\overline{X}_u+a}$ . Since $\omega$ is tracial, we have

\begin{equation*} m(\overline {X}_u+a)=\omega (v_av_a^*)=\omega (v_a^*v_a)=\omega (1)=1=m(\overline {X}_u).\end{equation*}

But, $\overline{X}_u+a \subset \overline{X}_u$ . Hence, $\overline{X}_u \backslash (\overline{X}_u+a)$ has measure zero. Hence, the set $X_u=\bigcup _{a \in P}\overline{X}_u \backslash (\overline{X}_u+a)$ has measure zero. Therefore, $m$ is concentrated on $F=\{G\}$ . Now the proof follows from Corollary 1.4 of [Reference Neshveyev22].

We end this section with a few definitions that we need later.

Example 2.1. The pair $(\overline{Y}_u,\overline{X}_u)$ is a $(G,P)$ -space. Define

\begin{equation*}Y_u\,:\!=\,\overline {Y}_u \backslash \{G\};\,\,X_u\,:\!=\, \overline {X}_u \backslash \{G\}.\end{equation*}

Then, $(Y_u,X_u)$ is pure.

We often abuse terminology and call an $e^{-\beta c}$ -conformal measure on a $(G,P)$ -space $(Y,X)$ an $e^{-\beta c}$ -conformal measure on $Y$ .

3. KMS states and conformal measures

Let us recall the definition of a $\beta$ -KMS state. Let $A$ be a $C^*$ -algebra, and suppose $\tau \,:\!=\{\tau _t\}_{t \in \mathbb{R}}$ is a $1$ -parameter group of automorphisms of $A$ . Suppose $\beta \in \mathbb{R}$ . A state $\omega$ on $A$ is a $\beta$ -KMS state for $\tau$ if

\begin{equation*}\omega(ab)=\omega(b\tau_{i\beta}(a)),\end{equation*}

for all $a, b$ in a norm dense $\tau$ -invariant $^*$ -algebra of analytic elements in $A$ .

Let $\omega$ be a $\beta$ -KMS state on $A$ . Then, $\omega$ is said to be extremal if it is an extreme point in the simplex of $\beta$ -KMS states. It is well known that a $\beta$ -KMS state $\omega$ is extremal if and only if the associated GNS representation $\pi _\omega$ is factorial, i.e. $\pi _\omega (A)^{\prime\prime}$ is a factor. We say an extremal $\beta$ -KMS state $\omega$ is of type $t$ if $\pi _\omega (A)^{\prime\prime}$ is a factor of type $t$ .

Fix a non-zero homomorphism $c\,:\,G \to \mathbb{R}$ . Recall that the $1$ -parameter group of automorphisms $\sigma ^{c}\,:\!=\{\sigma _{t}\}_{t \in \mathbb{R}}$ on $C_{c}^{*}(P)$ is defined by

\begin{equation*}\sigma_t(v_a)=e^{itc(a)}v_a,\end{equation*}

for $a \in P$ . For $t \in G$ , let $D(t), R(t)\subset \overline{X}_u$ be defined by

\begin{align*} D(t)&\,:\!=\{A\in \overline{X}_u\,:\, A+t \in \overline{X}_u \} \\ R(t)&\,:\!=\{A\in \overline{X}_u\,:\, A-t \in \overline{X}_u \}. \end{align*}

For $t \in G$ , denote the map $D(t) \ni A\to A+t \in R(t)$ by $T_t$ .

Let $\beta$ be a real number. As we have already described the tracial states on $C_c^{*}(P)$ in Proposition 2.1, we assume for the rest of this section that $\beta$ is an arbitrary, but a fixed non-zero real number. Let $\omega$ be a $\beta$ -KMS state on the $C^{*}$ -algebra $C_c^{*}(P)=C^*(\overline{X}_u \rtimes P)$ for $\sigma ^c$ . Restriction of $\omega$ to $C(\overline{X}_u)$ defines a probability measure $m$ on $\overline{X}_u$ . It can be checked from the covariance relation and the KMS condition that

\begin{equation*}\omega(f)=e^{-\beta c(t)}\omega(f\circ T_t),\end{equation*}

for every $f \in C(R(t))$ , or equivalently $m(E+t)=e^{-\beta c(t)}m(E)$ for every Borel subset $E$ of $ D(t)$ .

For $s,t \in G$ , we say $s \leq t$ if $t-s \in P$ . Then, $G$ with the preorder $\leq$ is a directed set. Note that $P$ is a cofinal subset of $G$ . Moreover, if $s \leq t$ , then $\overline{X}_u-s\subseteq \overline{X}_u-t$ . Let $s \in G$ . Define a measure $m_s$ on $\overline{X}_u-s$ by

\begin{equation*}m_s(E)=e^{\beta c(s)}m(E+s).\end{equation*}

Thanks to the fact that $m$ is conformal, it is clear that $m_t|_{\overline{X}_u-s}=m_s$ for $s\leq t$ . Define a $\sigma$ -finite measure $m_\omega$ on $\overline{Y}_u$ by setting

(3.4) \begin{equation} m_\omega (E)\,:\!=\, \lim _{s \in G}m_s(E \cap (\overline{X}_u-s))=\lim _{s \in P}m_s(E\cap (\overline{X}_u-s)). \end{equation}

Proof. It is clear that $m_\omega |_{\overline{X}_u}=m$ and so $m_\omega (\overline{X}_u)=1$ . For $t\in G$ and a Borel subset $E\subset \overline{Y}_u$ ,

\begin{eqnarray*} m_\omega (E+t)&=&\lim _sm_s((E+t)\cap (\overline{X}_u-s))\\ &=&\lim _s e^{\beta c(s)}m(((E+t)\cap (\overline{X}_u-s))+s)\\ &=& \lim _s e^{\beta c(s)}m((E\cap ( \overline{X}_u-(s+t)))+s+t)\\ &=& e^{-\beta c(t)}\lim _r e^{\beta c(r)} m((E\cap (\overline{X}_u-r))+r)\,\,(\textrm{by a change of variable} r=s+t)\\ &=& e^{-\beta c(t)}\lim _r m_r(E\cap (\overline{X}_u-r))\\ &=& e^{-\beta c(t)}m_\omega (E). \end{eqnarray*}

Hence the proof.

Recall that for an étale groupoid $\mathcal{G}$ , the map $f \mapsto f|_{\mathcal{G}^{(0)}}$ from $C_c(\mathcal{G})$ to $C_0(\mathcal{G}^{(0)})$ defines a conditional expectation $E\,:\,C^*(\mathcal{G}) \to C_0(\mathcal{G}^{(0)})$ . For $\mathcal{G}\,:\!=\,\overline{X}_u \rtimes P$ , the conditional expectation $E$ has the following form

\begin{equation*} E(fw_s)\,:\!=\,\begin {cases} f & \mbox{if} \,\, s=0,\\ 0 & \mbox { otherwise}. \end {cases} \end{equation*}

The next proposition is a direct consequence of Theorem 1.3 of [Reference Neshveyev22]. Hence, we omit the proof.

Proposition 3.2. The map $\omega \mapsto m_{\omega }$ from the set of $\beta$ -KMS states for $\sigma =\sigma ^c$ to the set of $e^{-\beta c}$ -conformal measures on $\overline{Y}_u$ is surjective. More specifically, for an $e^{-\beta c}$ -conformal $m$ on $\overline{Y}_u$ , the state $\omega _m$ on $C^{*}(\overline{X}_u \rtimes P)$ defined by

\begin{equation*}\omega _m(a) = \int _{\overline {X}_u} E(a)\,dm \end{equation*}

is a $\beta$ -KMS state such that $m_{\omega _{m}}=m$ .

Remark 3.1. In general, the map $\omega \to m_\omega$ of Proposition 3.1 need not be injective. The structure of a generic KMS state is determined by Neshveyev’s result (Theorem 1.3 of Reference Neshveyev[22] ). Let $\mathcal{G}$ be an étale groupoid, and let $\overline{c}\,:\,\mathcal{G} \to \mathbb{R}$ be a continuous homomorphism. For $t \in \mathbb{R}$ , let $\sigma _t$ be the automorphism of $C^{*}(\mathcal{G})$ defined by

\begin{equation*} \sigma _t(f)(\gamma )\,:\!=\,e^{it\overline {c}(\gamma )}f(\gamma )\end{equation*}

for $f \in C_{c}(\mathcal{G})$ . Then, $\sigma ^{\overline{c}}\,:\!=\{\sigma _t\}_{t \in \mathbb{R}}$ defines a $1$ -parameter group of automorphisms on $C^{*}(\mathcal{G})$ . Theorem 1.3 of Reference Neshveyev[22] describes the set of $\beta$ -KMS states for $\sigma ^{\overline{c}}$ .

In our situation, the groupoid $\mathcal{G}$ is the Deaconu-Renault groupoid $\overline{X}_u \rtimes P$ , and the homomorphism $\overline{c}\,:\,\mathcal{G} \to \mathbb{R}$ is given by

\begin{equation*} \overline {c}(A,g)=-c(g).\end{equation*}

One immediate corollary of Theorem 1.3 of Reference Neshveyev[22] is that, if the homomorphism $c\,:\,G \to \mathbb{R}$ is injective, then the map $\omega \to m_{\omega }$ of Proposition 3.1 is a bijection.

Proof. The proof is an adaptation of the proof of Lemma 3.6 in [Reference Christensen and Thomsen5] with appropriate modifications. Assume that $m_\omega$ is not ergodic. Then, there exists a $G$ -invariant Borel subset $A\subset \overline{Y}_u$ such that neither $A$ nor $A^c$ has measure zero. We claim that $0\lt m_\omega (A\cap \overline{X}_u)\lt 1$ and $0\lt m_\omega (A^c\cap \overline{X}_u)\lt 1$ .

Suppose that $m_\omega (A\cap \overline{X}_u)=0$ . By conformality, $m_\omega ((A\cap \overline{X}_u)-t)=0$ for every $t \in G$ . Hence, $\bigcup _{t \in G}(A\cap \overline{X}_u)-t$ is a null set. The fact that $A$ is $G$ -invariant implies that

\begin{equation*}\bigcup _{t \in G}(A\cap \overline {X}_u)-t=A \cap \big (\bigcup _{t \in G}(\overline {X}_u-t)\big )=A.\end{equation*}

Therefore, $m_\omega (A)=0$ which is a contradiction. This proves that $m_\omega (A\cap \overline{X}_u)\neq 0$ . Similarly, we can prove that $m_\omega (A^c \cap \overline{X}_u) \neq 0$ . Consequently, $0 \lt m_\omega (A \cap \overline{X}_u)\lt 1$ and $0 \lt m_\omega (A^c \cap \overline{X}_u)\lt 1$ . This proves the claim.

Let $(H_\omega,\pi _\omega, \Omega _\omega )$ be the GNS representation of $\omega$ . Extend the homomorphism $\pi _\omega |_{C(\overline{X}_u)}$ to a $^*$ - homomorphism $\overline{\pi }_\omega \,:\, L^{\infty }(\overline{X}_u,m_\omega ) \to \pi _\omega (C^*(\overline{X}_u \rtimes P))^{\prime\prime}$ by defining

\begin{equation*}\overline {\pi }_\omega (h)\,:\!=\,\lim _{ n \to \infty }\pi _\omega (f_n)\end{equation*}

Here, the limit is taken in the SOT sense and $\{f_n\}$ is any sequence in $C(\overline{X}_u)$ such that for every $n$ , $|f_n| \leq ||h||_\infty$ a.e and the sequence $(f_n(x)) \to h(x)$ for almost all $x$ .

For $s \in G$ , set $W_s=\pi _\omega (w_s)$ . Then, the following covariance relation is satisfied

\begin{equation*} W_s\overline {\pi }_\omega (f) W_s^*=\overline {\pi }_\omega (R_s(f))\end{equation*}

for $s \in G$ and $f \in L^{\infty }(\overline{X}_u,m_\omega )$ .

Next, we claim that $\overline{\pi }_\omega (1_{A \cap \overline{X}_u})$ and $\overline{\pi }_\omega (1_{A^c \cap \overline{X}_u})$ are central in $\pi _\omega (C^*(\overline{X}_u \rtimes P))^{\prime\prime}$ . We will give details for $\overline{\pi }_\omega (1_{A \cap \overline{X}_u})$ and the centrality of $\overline{\pi }_\omega (1_{A^c \cap \overline{X}_u})$ follows similarly. Let $s \in P$ be given. Note that $R_{-s}(1_{A \cap \overline{X}_u})=1_{(A-s) \cap (\overline{X}_u-s)\cap \overline{X}_u}=1_{A \cap \overline{X}_u}$ and $\overline{\pi }_\omega (f)$ commutes with $W_sW_s^*$ for every $f \in L^{\infty }(\overline{X}_u,m_\omega )$ .

Calculate as follows to observe that

\begin{eqnarray*} \overline{\pi }_\omega (1_{A \cap \overline{X}_u})W_s&=& \overline{\pi }_\omega (1_{A \cap \overline{X}_u})W_sW_s^*W_s\\ &=& W_sW_s^*\overline{\pi }_\omega (1_{A \cap \overline{X}_u})W_s\\ &=&W_s \overline{\pi }_\omega (R_{-s}(1_{A \cap \overline{X}_u}))\\ &=&W_s \overline{\pi }_\omega (1_{A \cap \overline{X}_u}). \end{eqnarray*}

Since $\{W_s\,:\, s \in P\}$ generates $(\pi _\omega (C^*(\overline{X}_u \rtimes P))^{\prime\prime}$ , we can conclude that $\pi _\omega (1_A \cap \overline{X}_u)$ is central. This proves the claim.

Define states $\omega _1$ and $\omega _2$ on $C^{*}(\overline{X}_u \rtimes P)$ by

\begin{align*} \omega _1(a)&=\frac{1}{m_\omega (A\cap \overline{X}_u)}\langle \overline{\pi }_\omega (1_{A \cap \overline{X}_u})\pi _\omega (a)\Omega _\omega,\Omega _\omega \rangle, \\ \omega _2(a)&=\frac{1}{m_\omega (A^c\cap \overline{X}_u)}\langle \overline{\pi }_\omega (1_{A^c \cap \overline{X}_u})\pi _\omega (a)\Omega _\omega,\Omega _\omega \rangle. \end{align*}

The centrality of the projections $\overline{\pi _\omega }(1_{A \cap \overline{X}_u})$ and $\overline{\pi _\omega }(1_{A^c \cap \overline{X}_u})$ imply that $\omega _1$ and $\omega _2$ are $\beta$ -KMS states on $C^*(\overline{X}_u \rtimes P)$ . Moreover, $\omega =m_\omega (A\cap \overline{X}_u)\omega _1+m_\omega (A^c\cap \overline{X}_u)\omega _2$ which contradicts the extremality of $\omega$ . Hence the proof.

Next, we work out the GNS representation of the KMS state $\omega _m$ given by Proposition 3.2. Suppose $m$ is an $e^{-\beta c}$ conformal measure on $\overline{Y}_u$ , and let $\omega _m$ be the $\beta$ -KMS state on the $C^*$ -algebra $C_{c}^{*}(P)=C^*(\overline{X}_u \rtimes P)$ obtained via the conditional expectation as in Proposition 3.2. Our goal is to show that if $(H_\omega,\pi _\omega, \Omega _\omega )$ is the GNS representation of $\omega _m$ , then the von Neumann algebra $(\pi _\omega (C^*(\overline{X}_u \rtimes P)))^{\prime\prime}$ is isomorphic to the full corner $1_{\overline{X}_u} (L^\infty (\overline{Y}_u)\rtimes G)1_{\overline{X}_u}$ of $L^\infty (\overline{Y}_u)\rtimes G$ .

Let $\lambda \,:\!=\{\lambda _s\}_{s \in G}$ be the Koopman representation of $G$ on $L^{2}(\overline{Y}_u,m)$ . Recall that

\begin{equation*}\lambda_s\xi(A)=e^\frac{\beta c(s)}2\xi(A-s),\end{equation*}

for $s \in G$ and $\xi \in L^{2}(\overline{Y}_u,m)$ .

Let $\mathcal{K}=l^2(G,L^2(\overline{Y}_u))$ and $\pi _0$ be the representation of $C_0(\overline{Y}_u)\rtimes G$ on $\mathcal{K}$ defined by

\begin{eqnarray*} \pi _0(fu_t)(\xi )(s)=f\lambda _t(\xi (s-t)). \end{eqnarray*}

Then, $(\pi _0(C_0(\overline{Y}_u) \rtimes G))^{\prime\prime}=L^\infty (\overline{Y}_u)\rtimes G$ . Let $\mathcal{H}$ be the Hilbert subspace of $\mathcal{K}$ given by

\begin{equation*}\mathcal {H}\,:\!=\{\xi \in l^2(G,L^2(\overline {Y}_u))\,:\, \xi (t)\in L^2(\overline {X}_u+t)\}.\end{equation*}

It is clear that $L^\infty (\overline{Y}_u)\rtimes G$ leaves $\mathcal{H}$ invariant, thus giving a normal representation $\overline{\pi }$ of the crossed product $L^\infty (\overline{Y}_u)\rtimes G$ on $\mathcal{H}$ . Let $\pi =\overline{\pi }\circ \pi _0$ .

Notation: For $s \in G$ and $\xi \in L^{2}(\overline{X}_u+s)$ , let $\xi \otimes \delta _s \in \mathcal{H}$ be defined by

\begin{equation*} \xi \otimes \delta _s(t)=\begin {cases} \xi & \mbox{if} \,\, t=s,\\ 0 & \mbox { otherwise}. \end {cases} \end{equation*}

Proof. Note that since $\pi =\overline{\pi }\circ \pi _0$ and the von Neumann algebra generated by $\pi _0(C_0(\overline{Y}_u) \rtimes G)$ is $L^\infty (\overline{Y}_u)\rtimes G$ , to complete the proof, it is enough to prove that $\overline{\pi }$ is injective.

First, we claim that $\overline{\pi }$ restricted to $L^{\infty }(\overline{Y}_u)$ is injective. Let $f \in L^{\infty }(\overline{Y}_u)$ be such that $\overline{\pi }(f)=0$ . Suppose $s \in G$ . Then, for every $\xi \in L^{2}(\overline{X}_u+s)$ ,

\begin{equation*} 0=\overline {\pi }(f)(\xi \otimes \delta _s)(s)=f\xi .\end{equation*}

Thus, $f=0$ a.e on $\overline{X}_u+s$ for every $s\in G$ . Since $\overline{Y}_u=\bigcup _{s \in G} \overline{X}_u+s$ , it follows that $f=0$ a.e on $\overline{Y}_u$ proving that $\overline{\pi }$ is faithful on $L^\infty (\overline{Y}_u)$ .

Let $E$ be the usual conditional expectation from $L^\infty (\overline{Y}_u)\rtimes G\to L^\infty (\overline{Y}_u)$ given by $ E(fu_s)=\delta _{s,0}f$ . Here, $\delta _{s,0}$ is the Kronecker delta. Then, the conditional expectation $E$ is faithful. Decompose $\mathcal{H}$ as $\displaystyle \mathcal{H}=\bigoplus _{t \in G}L^2(\overline{X}_u+t)$ , and for $ t \in G$ , let $P_t$ be the projection onto the subspace $L^2(\overline{X}_u+t)$ . Define a map $\widetilde{E}\,:\, B(\mathcal{H}) \to B(\mathcal{H})$ by

\begin{equation*}\widetilde E(T)=\sum_{t\in G}P_tTP_t,\end{equation*}

where the convergent sum is w.r.t. the strong operator topology. Clearly, $\widetilde{E}\circ \overline{\pi }=\overline{\pi }\circ E$ .

Let $x \in L^{\infty }(\overline{Y}_u) \rtimes G$ be such that $\overline{\pi }(x)=0$ . Then, $\overline{\pi }(x^*x)=0$ . Therefore,

\begin{equation*} \overline {\pi }(E(x^*x))=\widetilde {E}(\overline {\pi }(x^*x))=0.\end{equation*}

Since $\overline{\pi }$ is faithful on $L^{\infty }(\overline{Y}_u)$ , we have $E(x^*x)=0$ . Since $E$ is faithful, $x=0$ . This completes the proof.

Recall that $C^{*}(\overline{X}_u \rtimes P)$ is the full corner $p(C_0(\overline{Y}_u) \rtimes G)p$ where $p=1_{\overline{X}_u}$ . Consider the Hilbert space

\begin{equation*}P\mathcal {H}=\{\xi \in l^2(G,L^2(\overline {Y}_u))\,:\, \xi (t)\in L^2(\overline {X}_u\cap ( \overline {X}_u+t))\}\end{equation*}

where $P=\pi (p)$ . Define $\widetilde{\pi }\,:\, C^{*}(\overline{X}_u \rtimes P) \to B(P\mathcal{H})$ by $\widetilde{\pi }(fw_t)=P\pi (\tilde{f}u_t)P$ where $\tilde{f}=f1_{\overline{X}_u}$ . Define $\xi \in P\mathcal{H}$ by,

\begin{equation*} \xi (t)=\begin {cases} 1_{\overline {X}_u} & \mbox{if} \,\, t=0,\\ 0 & \mbox { otherwise}. \end {cases} \end{equation*}

Proof. Observe that, for $f \in C(\overline{X_u})$ , and $s,t\in G$ ,

\begin{align*} \widetilde{\pi}(fw_t)\xi(s)& =P\pi(\tilde f u_t)P\xi(s)\\ & =\pi(1_{\overline{X}_u}\tilde f u_t )\xi(s)\\ & = 1_{\overline{X}_u}\tilde f \lambda_t(\xi(s-t))\\ & = \delta_{s,t}1_{\overline{X}_u}f \lambda_t(\xi(0))\\ & =e^{\frac{\beta c(s)}{2}}\delta_{s,t} 1_{\overline{X}_u}f1_{\overline{X}_u+s}\end{align*}

From the above formula, it is clear that $\langle \widetilde{\pi }(a)\xi,\xi \rangle =\omega _m(a)$ for every $a \in C^{*}(\overline{X}_u \rtimes P)$ . Also, $span\{\widetilde{\pi }(fw_t)\xi \,:\, f \in C(\overline{X}_u), t \in G\}=span\{\eta \otimes \delta _s\,:\, \eta \in C(\overline{X}_u \cap (\overline{X}_u+s)), s \in G\}$ , and the latter set is dense in $P\mathcal{H}$ proving that $\xi$ is cyclic for $\widetilde{\pi }$ . Thus, $(\widetilde{\pi },P\mathcal{H},\xi )$ is the GNS representation of the $\beta$ -KMS state $\omega _m$ .

From the definition of $\widetilde{\pi }$ , we have

\begin{equation*}(\widetilde {\pi }(C^*(\overline {X}_u \rtimes P)))^{\prime\prime}=(P\pi (C_0(\overline {Y}_u) \rtimes G)P)^{\prime\prime}=P(\pi (C_0(\overline {Y}_u) \rtimes G))^{\prime\prime}P.\end{equation*}

The conclusion is now clear from Lemma 3.1.

We have the following main theorem of this section establishing the factor types of extremal $\beta$ -KMS states $\omega _m$ .

Proof. The first equivalence in $(2)$ follows from Lemma 3.2 and the following facts. It is well known that

1. (i) a $\beta$ -KMS state $\omega$ is extremal if and only if its GNS representation $\pi _\omega$ is factorial, and

2. (ii) a full corner $pMp$ is a factor of type $t$ if and only if $M$ is a factor of type $t$ .

Other conclusions are standard.

A few remarks are in order.

In the next proposition, we work out the extremal $\beta$ -KMS states whose associated measure is supported on an orbit.

Proof. The only thing that requires proof is that $\omega _{\chi,m}$ is of type I. Other assertions follow from Corollary 1.4 of [Reference Neshveyev22]. Let $m$ be an $e^{-\beta c}$ -conformal measure on $Y_u$ supported on an orbit $Orb(A)$ for some $A$ and denote the of $A$ by $H$ . Let $\chi$ be a character of $H$ , and let $\omega \,:\!=\,\omega _{\chi,m}$ . Let $\pi _\omega$ be the GNS representation of $\omega _{\chi,m}$ . To prove that $\omega _{\chi,m}$ is extremal of type I, it suffices to show that the von Neumann algebra $\pi _{\omega }(C^*(\overline{X}_u \rtimes P))^{\prime\prime}$ is a factor of type I. The proof is similar to the proof of Theorem 3.1.

By the conformality of $m$ , for a Borel set $E\subset Orb(A)$ with $m(E)\gt 0$ and $t \in H$ , we have

\begin{equation*}m(E)=m(E+t)=e^{-\beta c(t)}m(E),\end{equation*}

implying that $c(t)=0$ . Thus, $H\subset c^{-1}(0)$ . Let $\lambda =\{\lambda _t\}_{t \in G}$ be the Koopman representation of $G$ on $L^2(\overline{Y}_u)$ . Note that since $Y_u=Orb(A)$ upto a null set, $L^2(\overline{Y}_u)=L^2(Orb(A))$ and we can conclude that $\lambda _t=\lambda _s$ , whenever $t-s \in H$ . Extend the character $\chi$ of $H$ to a character of $G$ which we denote again by $\chi$ .

Let $\mathcal{K}$ be the Hilbert space defined by $\mathcal{K}=l^2(G/H,L^2(\overline{Y}_u))$ , and let $\pi _0$ be the representation of $C_0(\overline{Y}_u)\rtimes G$ on $\mathcal{K}$ given by

\begin{equation*}\pi_0(fu_t)(\xi)(\overline{s})=\overline{\chi(t)} f\lambda_{t}(\xi(\overline{s-t})).\end{equation*}

Clearly, $\pi _0(C_0(\overline{Y}_u)\rtimes G)^{\prime\prime}=L^\infty (\overline{Y}_u)\rtimes G/H$ . Let $\mathcal{H}$ be the Hilbert subspace of $\mathcal{K}$ given by

\begin{equation*}\mathcal {H}\,:\!=\{\xi \in l^2(G/H,L^2(\overline {Y}_u))\,:\, \xi (\overline {t})\in L^2(\overline {X}_u+t)\}.\end{equation*}

We have $\overline{s}=\overline{t}$ implies $\overline{X}_u+s=\overline{X}_u+t$ upto a null set, and hence, the definition of $\mathcal{H}$ makes sense. Since $L^\infty (\overline{Y}_u)\rtimes G/H$ leaves $\mathcal{H}$ invariant, we get a representation $\overline{\pi }\,:\, L^\infty (\overline{Y}_u)\rtimes G/H \to B(\mathcal{H})$ . Define $\pi =\overline{\pi }\circ \pi _0$ . Consider the Hilbert space $P\mathcal{H}$ where $P=\pi (1_{\overline{X}_u})$ . Let $\xi \in P\mathcal{H}$ be given by,

\begin{equation*} \xi (\overline {t})=\begin {cases} 1_{\overline {X}_u} & \mbox{if} \,\, \overline {t}=0,\\ 0 & \mbox { otherwise }. \end {cases} \end{equation*}

Let $\widetilde{\pi }\,:\, C^{*}(\overline{X}_u \rtimes P) \to B(P\mathcal{H})$ be defined by $\widetilde{\pi }(fw_t)=P\pi (\tilde{f}u_t)P$ where $\tilde{f}=f1_{\overline{X}_u}$ .

Arguing as before, we can conclude the following.

1. 1. The map $\overline{\pi }$ is injective, and hence, the von Neumann algebra $\pi (C_0(\overline{Y}_u)\rtimes G)^{\prime\prime}$ is isomorphic to $L^\infty (\overline{Y}_u)\rtimes G/H$ .

2. 2. The triple $(\widetilde{\pi },P\mathcal{H},\xi )$ is the GNS representation of $\omega _{\chi,m}$ .

3. 3. The von Neumann algebra generated by $\widetilde{\pi }(C^*(\overline{X}_u \rtimes P))$ is isomorphic to the full corner $1_{\overline{X}_u} (L^\infty (\overline{Y}_u)\rtimes G/H)1_{\overline{X}_u}$ .

Thus, $\pi _{\omega }(C^*(\overline{X}_u \rtimes P))^{\prime\prime}$ is isomorphic to the full corner $1_{\overline{X}_u} (L^\infty (\overline{Y}_u)\rtimes G/H)1_{\overline{X}_u}$ which is clearly a factor of type I (as the measure $m$ is atomic). This completes the proof.

We end this section with a proposition that allows us to construct conformal measures on the $(G,P)$ -space $(Y_u,X_u)$ , and consequently, KMS states on $C_c^{*}(P)$ of the desired type. Let $(Y,X)$ be a pure $(G,P)$ -space. For $y \in Y$ , let

\begin{equation*}Q_y\,:\!=\{s \in G\,:\, y-s \in X\}.\end{equation*}

First, let us check that for $y \in Y$ , $Q_y \in Y_u$ . Fix $y \in Y$ .

1. (1) Since, $\displaystyle Y=\bigcup _{a \in P}(X-a)$ , it follows that there exists $a \in P$ such that $y \in X-a$ . Then, $-a \in Q_y$ . Thus, $Q_y$ is non-empty.

2. (2) As the intersection $\displaystyle \bigcap _{a \in P}(X+a)=\emptyset$ , it follows that there exists $a \in P$ such that $y \notin X+a$ . Then, $a \notin Q_y$ for such an element $a$ . Hence, $Q_y$ is a proper subset of $G$ .

3. (3) The fact that $X+P \subset X$ implies that $-P+Q_y \subset Q_y$ .

Proposition 3.5. Let $(Y,X)$ be a pure $(G,P)$ -space, and let $m$ be an $e^{-\beta c}$ -conformal measure on $Y$ . Denote the map

\begin{equation*}Y \ni y \to Q_y \in Y_u,\end{equation*}

by $T$ . Assume that $T$ is $1$ - $1$ . Then, we have the following.

1. (1) The map $T$ is $G$ -equivariant, measurable, and $T^{-1}(X_u)=X$ .

2. (2) The push-forward measure $T_*m\,:\!=\,m \circ T^{-1}$ is an $e^{-\beta c}$ -conformal measure on the pure $(G,P)$ -space $(Y_u,X_u)$ . Moreover, $(Y,X,m)$ and $(Y_u,X_u,T_*m)$ are metrically isomorphic.

Consequently, for $t \in \{I,II,III\}$ , $(Y,m,G)$ is essentially free, ergodic and is of type $t$ if and only if $(Y_u,T_*m,G)$ is essentially free, ergodic and is of type $t$ .

Proof. Let $y \in Y$ and $s \in G$ be given. Note that for $t \in G$ , $t \in Q_{y+s}$ iff $y+s-t \in X$ iff $y-(t-s) \in X$ iff $t-s \in Q_y$ iff $t \in Q_y+s$ . Thus, $Q_{y+s}=Q_{y}+s$ . This shows that $T$ is $G$ -equivariant. To show that $T$ is measurable, it suffices to show that for every $s \in G$ , the map

\begin{equation*}Y \ni y \to 1_{Q_y}(s) \in \{0,1\}\end{equation*}

is measurable. Fix $s \in G$ . Clearly, for every $y \in Y$ ,

\begin{equation*} 1_{Q_y}(s)=1_{X+s} (y).\end{equation*}

Since $X+s$ is a measurable subset of $Y$ , it follows that the map

\begin{equation*}Y \ni y \to 1_{Q_y}(s) \in \{0,1\},\end{equation*}

is measurable. Hence, $T$ is a measurable map. Note that for $y \in Y$ , $y \in T^{-1}(X_u)$ if and only if $0 \in Q_y$ if and only if $y \in X$ . Thus, $T^{-1}(X_u)=X$ . This completes the proof of $(1)$ .

Since $T^{-1}(X_u)=X$ , and $m(X)=1$ , it follows that $T_*m(X_u)=1$ . Let $E$ be a Borel subset of $Y_u$ , and let $s \in G$ be given. Thanks to the equivariance of $T$ , we have $T^{-1}(E+s)=T^{-1}(E)+s$ . Since $m$ is $e^{-\beta c}$ -conformal,

\begin{equation*}T_*m(E+s)=m(T^{-1}(E+s))=m(T^{-1}(E)+s)=e^{-\beta c(s)}m(T^{-1}(E))=e^{-\beta c(s)}T_*m(E).\end{equation*}

Therefore, $T_*m$ is an $e^{-\beta c}$ -conformal measure on $(Y_u,X_u)$ .

Set $Y^{\prime}_u\,:\!=\,T(Y)$ , and $X^{\prime}_u\,:\!=\,Y^{\prime}_u\cap X_u$ . Since $T$ is one-one, and $Y$ and $Y_{u}$ are standard Borel spaces, $Y^{\prime}_u$ is a Borel subset of $Y_u$ , and the map $T\,:\,Y \to Y^{\prime}_u$ is a Borel isomorphism. Also, $T(X)=X^{\prime}_u$ . Observe that $T_*m$ is supported on $Y^{\prime}_u$ . Thus, $T$ is an isomorphism between $(Y,X,m)$ and $(Y^{\prime}_u,X^{\prime}_u,T_*m)$ . Since $Y^{\prime}_{u}$ is a co-null $G$ -invariant subset of $Y_u$ , $(Y^{\prime}_{u},X^{\prime}_{u},T_*m)$ and $(Y_u,X_u,T_*m)$ are metrically isomorphic. Hence, $(Y,X,m)$ and $(Y_u,X_u,T_*m)$ are metrically isomorphic. This completes the proof of $(2)$ .

The final assertion is clear as metric isomorphism preserves essential freeness, ergodicity and type.

4. The case $P=\mathbb{N}^{2}$

In this section, we discuss the structure of KMS states on $C_{c}^{*}(P)$ for $\sigma ^{c}$ when $P=\mathbb{N}^2$ . Let us fix notation. Set $e_1\,:\!=\,(1,0)$ and $e_2\,:\!=\,(0,1)$ . Define $v_1\,:\!=\,e_1$ and $v_2\,:\!=\,e_1+e_2$ . Let $c\,:\,\mathbb{Z}^2 \to \mathbb{R}$ be a non-zero homomorphism. We normalise and assume that $c(e_1)=1$ and $c(e_2)=\theta$ . This does not lead to any loss of generality. In what follows, the homomorphism $c$ will be fixed.

First, we obtain a reasonable parametrisation of $Y_u$ . Let $\Omega \,:\!=\{0,1\}^{\mathbb{Z}}$ be the Cantor space. Consider the Bernoulli shift $\tau \,:\,\Omega \to \Omega$ defined by

\begin{equation*}\tau (x)_k\,:\!=\,x_{k-1}.\end{equation*}

Define a $\mathbb{Z}^2$ -action on $\Omega \times \mathbb{Z}$ by the following formulae.

\begin{align*} \big (x,t\big )+v_1&=\big (\tau (x),t+x_{-1}\big ),\text{ and }\\ \big (x,t\big )+v_2&=\big (x,t+1\big ). \end{align*}

For $(x,t)\in \Omega \times \mathbb{Z}$ , define $a(x,t)\,:\!=\,a=(a_m)_{m\in \mathbb{Z}}$ as follows.

(4.5) \begin{equation} a(x,t)_m\,:\!=\,a_m=\begin{cases} t-(x_0+x_1+\cdots +x_{m-1}) & \mbox{if} m\gt 0, \\ t & \mbox{if $m=0$}, \\ t+(x_{-1}+x_{-2}+\cdots +x_{m}) & \mbox{if} m\lt 0. \end{cases} \end{equation}

Note that $a_m-a_{m+1}=x_m$ . Let $A(x,t)$ be defined by

\begin{equation*}A(x,t)\,:\!=\{mv_1+nv_2\,:\, n \leq a_m\}.\end{equation*}

It is not difficult to verify that $-e_1+A(x,t) \subset A(x,t)$ and $-e_2+A(x,t) \subset A(x,t)$ . In other words, $A(x,t) \in Y_u$ for every $(x,t) \in \Omega \times \mathbb{Z}$ .

Proof. It is clear from equation (4.5) that $(1)$ and $(2)$ are equivalent. Suppose that $(2)$ holds. Let $m,n \in \mathbb{Z}$ . Suppose $1_A(mv_1+nv_2)=1$ . Then, $n \leq a_m$ . Since $a_m^{(k)} \to a_m$ , it follows that eventually $a_{m}^{(k)}=a_m$ . Thus, for large $k$ , $n \leq a_{m}^{(k)}$ , i.e. for large $k$ , $1_{A_k}(mv_1+nv_2)=1$ . If $1_{A}(mv_1+nv_2)=0$ . By the same argument, we can conclude that for large $k$ , $1_{A_k}(mv_1+nv_2)=0$ . Thus, $1_{A_k}(mv_1+nv_2) \to 1_{A}(mv_1+nv_2)$ for every $m,n \in \mathbb{Z}$ . Hence, $(A_k)_k \to A$ . This completes the proof of the implication $(2) \implies (3)$ .

Assume that $(3)$ holds. Let $m \in \mathbb{Z}$ be given. Note that $1_A(mv_1+a_mv_2)=1$ . Since $(A_k)_k \to A$ , it follows that for large $k$ , $1_{A_k}(mv_1+a_mv_2)=1$ . In other words, for large $k$ , $a_m \leq a_{m}^{(k)}$ . Also, note that $1_{A}(mv_1+(a_m+1)v_2)=0$ . Since $(A_k)_k \to A$ , it follows that for large $k$ , $1_{A_k}(mv_1+(a_m+1)v_2)=0$ , i.e. for large $k$ , $a_m+1\gt a_{m}^{(k)}$ . Thus, eventually $a_m\leq a_{m}^{(k)}\lt a_m+1$ , i.e. eventually $a_{m}^{(k)}=a_m$ . Hence, for every $m \in \mathbb{Z}$ , $a_{m}^{(k)} \to a_m$ as $k \to \infty$ . This completes the proof of the implication $(3) \implies (2)$ .

Proposition 4.1. With the foregoing notation, the map

\begin{equation*}\Omega \times \mathbb {Z}\ni (x,t)\mapsto A(x,t)\in Y_u\end{equation*}

is a $\mathbb{Z}^2$ -equivariant homeomorphism.

Proof. First, we check that the prescribed map is $\mathbb{Z}^2$ -equivariant. Let $(x,t) \in \Omega \times \mathbb{Z}$ be given. By definition, for $m,n \in \mathbb{Z}$ , $mv_1+nv_2 \in A(x,t)+v_1$ if and only if $n \leq a(x,t)_{m-1}$ . Thus,

\begin{equation*} A(x,t)+v_1=\{mv_1+nv_2\,:\, n \leq a(x,t)_{m-1}\}.\end{equation*}

From equation (4.5), it is clear that $a(\tau (x),t+x_{-1})_m=a(x,t)_{m-1}$ . Thus, $A(x,t)+v_1=A(\tau (x),t+x_{-1})$ .

For $m,n \in \mathbb{Z}$ , $mv_1+nv_2 \in A(x,t)+v_2$ if and only if $n \leq a(x,t)_m+1$ . Thus,

\begin{equation*} A(x,t)+v_2=\{mv_1+nv_2\,:\, n \leq a(x,t)_m+1\}.\end{equation*}

From equation (4.5), it is clear that $a(x,t)_m+1=a(x,t+1)_m$ . Therefore, $A(x,t)+v_2=A(x,t+1)$ . As a consequence, $A(x,t)+v_1=A(\tau (x),t+x_{-1})$ and $A(x,t)+v_2=A(x,t+1)$ . Since $\{v_1,v_2\}$ is a $\mathbb{Z}$ -basis for $\mathbb{Z}^2$ , it follows that the map

\begin{equation*}\Omega \times \mathbb{Z} \ni (x,t) \mapsto A(x,t) \in Y_u\end{equation*}

is $\mathbb{Z}^2$ -equivariant.

Let $(x,t), (y,s) \in \Omega \times \mathbb{Z}$ be given. Suppose $A(x,t)=A(y,s)$ . Then,

\begin{equation*} \{mv_1+nv_2\,:\, n \leq a(x,t)_m\}=\{mv_1+nv_2\,:\, n \leq a(y,s)_m\}.\end{equation*}

The above equality clearly implies that $a(x,t)_m=a(y,s)_m$ for every $m$ . It follows from equation (4.5) that $x=y$ and $t=s$ . Thus, the map $\Omega \times \mathbb{Z} \ni (x,t) \mapsto A(x,t) \in Y_u$ is $1$ - $1$ .

We now show that it is onto. Let $A \in Y_u$ be given. By translating, if necessary, we can assume that $0 \in A$ which implies that $-\mathbb{N}^2 \subset A$ .

Let $m \in \mathbb{Z}$ be given. We claim that the set $\{k \in \mathbb{Z}\,:\, mv_1+kv_2\in A\}$ is non-empty and bounded above. Choose $k_0\lt 0$ such that $m+k_0\lt 0$ . Then, we have

\begin{equation*} mv_1+k_0v_2=(m+k_0)e_1+k_0e_2\in -\mathbb {N}^2\subseteq A.\end{equation*}

Hence, the set $\{k \in \mathbb{Z}\,:\, mv_1+kv_2\in A\}$ is non-empty. Suppose $\{k \in \mathbb{Z}\,:\, mv_1+kv_2\in A\}$ is not bounded above. Then, for every $k\in \mathbb{N}$ , there exists a natural number $n_k\geq k$ such that $mv_1+n_kv_2\in A$ . Since $A-\mathbb{N}^2 \subset A$ , we have $mv_1+(n_k-\ell )v_2\in A$ for every $\ell \in \mathbb{N}$ and for every $k \in \mathbb{N}$ . This means that, for every $n \in \mathbb{Z}$ , $mv_1+nv_2\in A$ . Again for any $k,\ell \in \mathbb{N}$ and $n\in \mathbb{Z}$ , we have

\begin{equation*}(m+n-k)e_1+(n-\ell )e_2=mv_1+nv_2-ke_1-\ell e_2\in A-\mathbb {N}^2 \subset A.\end{equation*}

This implies that $A=\mathbb{Z}^2$ which contradicts the fact that $A \in Y_u$ . Hence, $\{k \in \mathbb{Z}\,:\, mv_1+kv_2\in A\}$ is bounded above. The proof of the claim is now complete.

For $m\in \mathbb{Z}$ , define

\begin{equation*}a_m\,:\!=\,\text {max}\{k \in \mathbb {Z}\,:\, mv_1+kv_2\in A\}.\end{equation*}

Then, $a_m$ is an integer. Let $A_a=\{mv_1+nv_2\,:\, n \leq a_m\}$ . We claim that $A=A_a$ and for every $m \in \mathbb{Z}$ , $0 \leq a_{m}-a_{m+1} \leq 1$ .

Clearly, $A \subset A_a$ . Suppose $mv_1+nv_2\in A_a$ . By the definition of $a_m$ , $mv_1+a_mv_2\in A$ . Note that

\begin{equation*} mv_1+nv_2 =mv_1+a_mv_2+(n-a_m)(e_1+e_2) \in A-\mathbb {N}^2\subseteq A.\end{equation*}

Hence, $A_a\subseteq A$ . Therefore, $A=A_a$ .

Let $mv_1+nv_2\in A_a$ be given. Since $A_a-\mathbb{N}^2 \subset A_{a}$ , we have $mv_1+nv_2-ke_1-\ell e_2\in A_a$ , for every $k,l\geq 0$ . Therefore, whenever $mv_1+nv_2\in A_a$ , we have $a_{m-k+\ell }\geq n-\ell$ . In particular, when $n=a_m,k=1$ and $\ell =0$ , we have $a_{m-1}\geq a_m$ , and when $n=a_m,k=0$ and $\ell =1$ , we have $a_{m+1}\geq a_m-1$ . Thus, $0\leq a_m-a_{m+1}\leq 1$ for every $m\in \mathbb{Z}$ . The proof of the claim is now over.

Define $x \in \Omega$ by setting $x_m\,:\!=\,a_{m}-a_{m+1}$ and let $t\,:\!=\,a_0$ . Clearly, for every $m$ , $a_m=a(x,t)_m$ . Therefore, $A_a=A(x,t)$ . Consequently, $A=A(x,t)$ . This proves the surjectivity of the map

\begin{equation*} \Omega \times \mathbb {Z} \ni (x,t) \to A(x,t) \in Y_u.\end{equation*}

The fact that the map $\Omega \times \mathbb{Z} \ni (x,t) \to A(x,t) \in Y_u$ is a homeomorphism follows from Lemma 4.1. This completes the proof.

Remark 4.1. For $A \in Y_u$ , let $G_A$ be the stabiliser of $A$ , i.e.

\begin{equation*}G_A\,:\!=\{(m,n) \in \mathbb {Z}^2\,:\, A+(m,n)=A\}.\end{equation*}

Let $(x,t) \in \Omega \times \mathbb{Z}$ , and let $A\,:\!=\,A(x,t) \in Y_u$ be the set defined as in Proposition 4.1 . Then, $G_A \neq 0$ if and only if $x$ is a periodic point, i.e. there exists $p\gt 0$ such that $x_{m+p}=x_m$ for all $m \in \mathbb{Z}$ . Moreover, there are only countably many periodic points in $\Omega$ . This has the consequence that the groupoid $\mathcal{G}\,:\!=\,\overline{X}_u \rtimes \mathbb{N}^2$ (and also the transformation groupoid $\overline{Y}_u \rtimes \mathbb{Z}^2$ ) has only countably many points in its unit space whose stabiliser is non-trivial.

We identify $Y_u$ with $\Omega \times \mathbb{Z}$ via the map prescribed in Proposition 4.1. After an abuse of notation, we write $Y_u=\Omega \times \mathbb{Z}$ . Then, $X_u=\Omega \times \mathbb{N}$ . Let $m$ be a probability measure on $\Omega$ . Define a measure $\overline{m}$ on $\Omega \times \mathbb{Z}$ by setting

\begin{equation*}\overline{m}(E\times \{n\})=(1-e^{-\beta(1+\theta)})e^{-\beta (1+\theta)n}m(E)\end{equation*}

for a measurable subset $E \subset \Omega$ .

Define $\chi :\Omega \to \mathbb{R}$ by

\begin{equation*} \chi (x)\,:\!=\,\begin {cases} 1 & \mbox { if } x_{-1} =0,\\ - \theta & \mbox { if } x_{-1}=1. \end {cases} \end{equation*}

Let $\beta$ be a real number, and let $m$ be a probability measure on $\Omega$ . The probability measure $m$ is said to be $e^{-\beta \chi }$ -conformal for the Bernoulli shift $\tau$ if for every Borel subset $E \subset \Omega$ ,

\begin{equation*} m(\tau (E))=\int _{E} e^{-\beta \chi }dm.\end{equation*}

Proof. The proof is not difficult. We have included some details for completeness. Suppose that $m$ is an $e^{-\beta \chi }$ -conformal measure on $\Omega$ . Clearly, $\overline{m}(X_u)=1$ . Let $F \subset \Omega$ be a Borel subset and let $n \in \mathbb{Z}$ be given. It suffices to show that

\begin{align*} \overline{m}((F\times \{n\})+v_1)&=e^{-\beta }\overline{m}(F\times \{n\}),\,\textrm{ and }\\ \overline{m}((F\times \{n\})+v_2)&=e^{-\beta (1+\theta )}\overline{m}(F\times \{n\}). \end{align*}

Define $F_0\,:\!=\{x \in F\,:\, x_{-1}=0\}$ and $F_1\,:\!=\{x \in F\,:\, x_{-1}=1\}$ . Calculate as follows to observe that

\begin{align*} \frac{1}{1-e^{-\beta (1+\theta )}}\overline{m}((F\times \{n\})+v_1) &=\frac{1}{1-e^{-\beta (1+\theta )}}\Big (\overline{m}(\tau (F_0)\times \{n\})+\overline{m}(\tau (F_1)\times \{n+1\})\Big )\\ &=e^{-\beta n(1+\theta )} m(\tau (F_0))+ e^{-\beta (n+1)(1+\theta )}m(\tau (F_1))\\ &=e^{-\beta n(1+\theta )} e^{-\beta } m(F_0)+ e^{-\beta (n+1)(1+\theta )} e^{\beta \theta }m(F_1)\\ &=e^{-\beta n(1+\theta )} e^{-\beta }\big (m(F_0)+ m(F_1)\big )\\ &=e^{-\beta }\frac{1}{1-e^{-\beta (1+\theta )}}\overline{m}(F\times \{n\}). \end{align*}

Similarly, we can prove that $ \overline{m}((F\times \{n\})+v_2)=e^{-\beta (1+\theta )}\overline{m}(F\times \{n\})$ . Hence, $\overline{m}$ is an $e^{-\beta c}$ -conformal measure on $(\Omega \times \mathbb{Z},\Omega \times \mathbb{N})$ .

Conversely, suppose $\mu$ is an $e^{-\beta c}$ -conformal measure on $(Y_u,X_u)$ . Define a measure $m$ on $\Omega$ by

\begin{equation*}m(E)=\frac1{1-e^{-\beta(1+\theta)}}\mu(E\times\{0\}),\end{equation*}

for a Borel subset $E \subset \Omega$ . Note that for a Borel set $E \subset \Omega$ , $E \times \{n\}=(E \times \{0\})+nv_2$ . Using the conformality condition on $\mu$ and the fact that $E \times \{n\}=(E \times \{0\})+nv_2$ , it is routine to see that $\overline{m}=\mu$ .

Let $F \subset \Omega$ be measurable. Set $F_0\,:\!=\{x\in F\,:\, x_{-1}=0\}$ and $F_1\,:\!=\{x\in F\,:\, x_{-1}=1\}$ . Thanks, to the conformality condition on $\mu$ and the calculation done earlier, it follows that the equality

\begin{equation*}\mu((F\times\{0\})+v_1)=e^{-\beta}\mu(F\times\{0\}),\end{equation*}

is equivalent to the equality

\begin{equation*} m(\tau (F_0))+ e^{-\beta (1+\theta )}m(\tau (F_1))= e^{-\beta }\big (m(F_0)+ m(F_1)\big )=e^{-\beta }m(F).\end{equation*}

In other words, for every Borel subset $F \subset \Omega$ ,

\begin{equation*} m(F)=\int _{\tau (F)}e^{\beta (\chi \circ \tau ^{-1})}dm.\end{equation*}

Hence, $\frac{d(m\circ \tau )}{dm}=e^{-\beta \chi }$ , i.e. $m$ is $e^{-\beta \chi }$ -conformal. The fact that the map $m \to \overline{m}$ is a bijection is clear.

Remark 4.2. Note that for the existence of an $e^{-\beta c}$ -conformal measure on $(Y_u,X_u)$ , it is necessary that $\beta (1+\theta )\gt 0$ . This is because, if $\overline{m}$ is an $e^{-\beta c}$ -conformal measure on $Y_u$ , then

\begin{equation*} 1=\overline {m}(X_u)=\sum _{n=0}^{\infty }\overline {m}(\Omega \times \{n\})=\sum _{n=0}^{\infty }\overline {m}((\Omega \times \{0\})+nv_2)=\sum _{n=0}^{\infty }e^{-\beta (1+\theta )n}\overline {m}(\Omega \times \{0\}).\end{equation*}

Hence, the condition $\beta (1+\theta )\gt 0$ is necessary.

The bijection $m \to \overline{m}$ of Proposition 4.2 preserves essential freeness, ergodicity and type. Thus, $(\Omega,\tau,m)$ is essentially free if and only if $(Y_u,\mathbb{Z}^2,\overline{m})$ is essentially free. Similarly, $(\Omega,\tau,m)$ is ergodic and is of type $t$ if and only if $(Y_u,\mathbb{Z}^2,\overline{m})$ is ergodic and is of type $t$ . Here, $t \in \{I,II,III\}$ .

We are now in a position to determine the values of $\beta$ and $\theta$ for which there is a $\beta$ -KMS state on $C_{c}^{*}(\mathbb{N}^2)$ for $\sigma ^{c}\,:\!=\{\sigma _t\}_{t \in \mathbb{R}}$ . Recall that the flow $\sigma ^{c}\,:\!=\{\sigma _t\}_{t \in \mathbb{R}}$ on $C_{c}^{*}(\mathbb{N}^2)$ is defined by

\begin{equation*} \sigma _t(v_{(m,n)})=e^{i(m+n\theta )t}v_{(m,n)}.\end{equation*}

Proof. The equivalence between $(1)$ , $(2)$ and $(3)$ follow from Propositions 3.2, 4.2 and Remark 4.2. We prove the equivalence between $(3)$ and $(4)$ . Assume that $(3)$ holds. Let $m$ be an $e^{-\beta \chi }$ -conformal measure on $\Omega$ . Suppose $\theta \lt 0$ . Then, the potential $\beta \chi$ is strictly negative or strictly positive. Suppose $\beta \chi$ is strictly positive. Then, there exists a real number $a\gt 0$ such that $\beta \chi \gt a$ . Note that

\begin{equation*} 1=m(\tau (\Omega ))=\int _{\Omega }e^{-\beta \chi }dm \leq e^{- a}\lt 1.\end{equation*}

This is a contradiction. A similar contradiction will be met if $\beta \chi$ is strictly negative. Therefore, $\theta \geq 0$ . (A direct application of Theorem 6.2 of [Reference Christensen and Thomsen5] could also have been made instead). The condition $\beta (1+\theta )\gt 0$ implies that $\beta \gt 0$ . This completes the proof of $(3) \implies (4)$ .

Assume now that $\beta \gt 0$ and $\theta \geq 0$ . Let us fix some notation. For $n \in \mathbb{Z}$ , define $S_{n}(\chi )$ by setting

\begin{equation*} S_n(\chi )(x)\,:\!=\,\begin {cases} \sum _{j=0}^{n-1} \chi \circ \tau ^{j}(x) & \mbox { if } n \geq 1,\\ 0 & \mbox { if } n=0 \\ - \sum _{j=1}^{|n|}\chi \circ \tau ^{-j}(x) & \mbox { if } n \leq -1. \end {cases} \end{equation*}

Note that if $n \leq -1$ , then $S_{n}(\chi )(x)=-|n|+(1+\theta )(x_{0}+x_{1}+\cdots +x_{|n|-1})$ , and if $n \geq 1$ , we have $S_n(\chi )(x)=n-(1+\theta )(x_{-1}+x_{-2}+\cdots +x_{-n})$ .

We apply Lemma 4.3 of [Reference Christensen and Thomsen5] which states the following. Let $x\in \Omega$ be given. If $x$ is periodic of period $p\gt 0$ , then there is an $e^{-\beta \chi }$ -conformal measure on $\Omega$ concentrated on the orbit of $x$ iff $S_p(\chi )(x)=0$ . If $x$ is not periodic, then there is an $e^{-\beta \chi }$ -conformal measure concentrated on the orbit of $x$ if and only if $\displaystyle \sum _{n\in \mathbb{Z}} e^{-\beta S_n(\chi )(x)}\lt \infty$ .

Case 1: Suppose $\theta =0$ . Let $x\in \Omega$ be such that $x_m=1$ for every $m \in \mathbb{Z}$ . Clearly, $x$ is periodic of period $1$ , and $S_{1}(\chi )(x)=0$ . Therefore, there is an $e^{-\beta \chi }$ -conformal measure supported on the orbit of $x$ .

Case 2: Suppose $\theta \gt 0$ . We exhibit uncountably many non-periodic points $x$ for which the series $\displaystyle \sum _{n \in \mathbb{Z}}e^{-\beta S_n(\chi )(x)}$ converges. Let $\ell$ be a positive integer such that $\ell \gt 1+\theta$ . Suppose $S \subset \ell \mathbb{N}$ . Denote the indicator function of $S$ by $1_{S}$ . Define $x \in \Omega$ by setting

\begin{equation*} x_m\,:\!=\,\begin {cases} 1 & \mbox { if} \, m \geq 0 \\ 1_{S}({-}m) & \,\mbox {if} \, m\lt 0 \end {cases} \end{equation*}

Note that $S_{n}(\chi )(x)=|n|\theta$ if $n \leq -1$ . Therefore, the series $\sum _{n\leq -1}^{\infty }e^{-\beta S_n(\chi )(x)}$ is convergent.

For $n \geq 1$ , observe that since $x_{-1}+x_{-2}+\cdots +x_{-n} =|S \cap \{1,2,\cdots,n\}| \leq \frac{n}{\ell }$ ,

\begin{equation*} S_{n}(\chi )(x)=n-(1+\theta )(x_{-1}+x_{-2}+\cdots +x_{-n}) \geq n(1-\frac {1+\theta }{\ell }).\end{equation*}

Thus, $ e^{-\beta S_n(\chi )(x)} \leq e^{-\beta n (1-\frac{1+\theta }{\ell })}$ . Consequently, the series $\sum _{n \geq 1}^{\infty }e^{-\beta S_n(\chi )(x)}$ converges. Thanks to Lemma 4.9 of [Reference Christensen and Thomsen5], there is an $e^{-\beta \chi }$ -conformal measure supported on the orbit of $x$ . The proof is complete.

We discuss the structure of KMS states in more detail now. Assume hereafter that $\beta$ is an arbitrary positive real number.

The case $\theta =0$ : Assume that $\theta =0$ . Denote the element in $\{0,1\}^{\mathbb{Z}}$ whose every entry is 1 by $\underline{1}$ . Suppose that $m$ is an $e^{-\beta \chi }$ -conformal measure on $\Omega$ . We claim that $m$ is concentrated on $\underline{1}$ . We claim that $m(\{x\,:\,x_{-1}=0\})=0$ . Suppose $m(\{x\,:\,x_{-1}=0\})\gt 0$ . Then, by conformality, we have

\begin{align*} 1&=m(\tau (\Omega ))\\ &=\int e^{-\beta \chi }dm\\ &=e^{-\beta }m(\{x\,:\,x_{-1}=0\})+m(\{x\,:\,x_{-1}=1\})\\ &\lt m(\{x\,:\,x_{-1}=0\})+m(\{x\,:\,x_{-1}=1\})=1 \end{align*}

which is a contradiction.

Therefore, for almost all $x$ , $x_{-1}=1$ . Since $m$ is quasi-invariant for the Bernoulli shift $\tau$ , we have, for every $k$ , $x_k=1$ for almost all $x$ . Hence, $x=\underline{1}$ for almost all $x$ . This proves the claim.

Let $\omega$ be a $\beta$ -KMS state on $C_{c}^{*}(\mathbb{N}^2)$ . Thanks to Propositions 4.1 and 4.2 and by what we have proved now, its associated $e^{ - \beta c}$ -conformal measure $m_{\omega }$ is supported on $Orb(A(\underline{1},0))$ . Note that

\begin{equation*}A\,:\!=\,A(\underline{1},0)=\{(m,n)\,:\, m \leq 0\}\end{equation*}