Proof Suppose first that h is coarse. Since $\mathrm {C}^{*}_u[X]$ is dense in $\mathrm {C}^{*}_u(X)$ , it is enough to show that the map in (2.2) is continuous for each $a\in \mathrm {C}^{*}_u[X]$ . Moreover, since $\mathrm {C}^{*}_u[X]$ is spanned by the subset of all $av_f$ , for $a\in \ell _\infty (X)$ and $f\colon A\subseteq X\to B\subseteq X$ a partial translation, it is enough to notice that (2.2) holds for all such elements $av_f$ . Fix such a and f. Then, as f is a partial bijection, we have that

(2.3) $$ \begin{align}\|\sigma_{h,t}(av_f)-\sigma_{h,s}(av_f)\|&=\|e^{it\bar h}av_fe^{-it\bar h}-e^{is\bar h}av_fe^{-is\bar h}\|\\ &=\sup_{x\in A}\Big|e^{it (h(f(x))-h(x))}-e^{is (h(f(x))-h(x))}\Big|a_{f(x),f(x)}.\notag \end{align} $$

Since f is a partial translation and h is coarse, we have

$$\begin{align*}\sup_{x\in A}|h(f(x))-h(x)|<\infty.\end{align*}$$

Therefore, it follows from (2.3) and the intermediate value theorem that

$$\begin{align*}t\in \mathbb{R}\mapsto \sigma_t(av_f)\in \mathrm{C}^{*}_u(X)\end{align*}$$

is continuous.

Suppose now that the action $\sigma _h$ is strongly continuous. Suppose towards a contradiction that h is not coarse. Then there is $r>0$ , and sequences $(x_i)_i$ and $(y_i)_i$ in X such that $\lim _i|h(x_i)-h(y_i)|=\infty $ and $d(x_i,y_i)\leq r$ for all $i\in \mathbb {N}$ . As X is u.l.f., those sequences cannot be bounded, so, by going to a subsequence if necessary, we assume that $(x_i)_i$ and $(y_i)_i$ are sequences of distinct points of X. We can then define a map

$$ \begin{align*}f\colon \{x_i\mid i\in \mathbb{N}\}&\to \{y_i\mid i\in \mathbb{N}\}\\ x_i&\mapsto y_i \end{align*} $$

and this map is a partial translation. So, $v_f\in \mathrm {C}^{*}_u[X]$ and, since $\sigma _h$ is strongly continuous, we have that

$$\begin{align*}\lim_{t\to 0}\|\sigma_t(v_f)-v_f\|=0.\end{align*}$$

Fix $\delta>0$ such that

$$\begin{align*}|t|<\delta\ \text{ implies } \ \|\sigma_t(v_f)-v_f\|<2.\end{align*}$$

Notice now that

$$ \begin{align*} \|\sigma_t(v_f)-v_f\|&= \|e^{it\bar h}v_fe^{-it\bar h}-v_f\|\\ &=\sup_{x\in X}\Big|e^{it (h(f(x))- h(x))}-1\Big| \\ &\geq \sup_{i\in \mathbb{N}}\Big|e^{it (h(y_i)- h(x_i))}-1\Big|. \end{align*} $$

Hence, picking $i\in \mathbb {N}$ large enough so that

$$\begin{align*}t=\frac{\pi}{|h(y_i)-h(x_i)|}<\delta,\end{align*}$$

we obtain that $\|\sigma _t(v_f)-v_f\|\geq 2$ ; contradiction.

Proof We first notice that the condition of $\sigma \colon \mathbb {R}\curvearrowright \mathrm {C}^{*}_u(X)$ leaving $\ell _\infty (X)$ invariant implies that $\sigma _t$ is the identity on $\ell _\infty (X)$ for all $t\in \mathbb {R}$ . Indeed, as $\sigma _0$ is by hypothesis the identity on $\mathrm {C}^{*}_u(X)$ , we have that $\sigma _0(e_{x,x})=e_{x,x}$ for all $x\in X$ . As $\sigma _t$ is an isomorphism for all $t\in \mathbb {R}$ , $\sigma _t(e_{x,x})$ must be a projection for all $t\in \mathbb {R}$ and all $x\in X$ . Therefore, since $t\in \mathbb {R}\mapsto \sigma _t(e_{x,x})\in \ell _\infty (X)$ is continuous, this shows that $\sigma _t(e_{x,x})=e_{x,x}$ for all $t\in \mathbb {R}$ and all $x\in X$ . Hence, $\sigma _t$ must be the identity on $c_0(X)$ for all $t\in \mathbb {R}$ . As isomorphisms of uniform Roe algebras are strongly continuous [Reference Špakula and WillettŠW13, Lemma 3.1], this shows that each $\sigma _t$ is the identity on $\ell _\infty (X)$ are desired.

Fix $x\in X$ . For each $\xi \in \ell _2(X)$ , let $r_\xi $ be the rank one operator given by

$$\begin{align*}r_\xi\zeta=\langle \zeta,\delta_x\rangle\xi \ \text{ for all }\ \zeta\in \ell_2(X).\end{align*}$$

For each $t\in \mathbb {R}$ , define an operator $u_t$ on $\ell _2(X)$ by letting

$$\begin{align*}u_t\xi=\sigma_t(r_\xi)\delta_x\ \text{ for all }\ \xi\in \ell_2(X).\end{align*}$$

Claim 2.3 We have

$$\begin{align*}\sigma_t(a)=u_tau_{-t}\ \text{ for all} \ a\in \mathrm{C}^{*}_u(X)\ \text{ and all }\ t\in \mathbb{R}. \end{align*}$$

In particular, $u_t\in \ell _\infty (X)$ for all $t\in \mathbb {R}$ .

Proof First notice that

(2.4) $$ \begin{align} ae_{x,x}=r_{a\delta_x}\ \text{ for all }a\in \mathrm{C}^{*}_u(X). \end{align} $$

Hence, by the arbitrariness of a above, this implies that

$$\begin{align*}u_tau_{-t}\xi=u_ta\sigma_{-t}(r_\xi)\delta_x=\sigma_t(a\sigma_{-t}(r_\xi))\delta_x=\sigma_t(a)r_\xi\delta_x=\sigma_t(a)\xi\end{align*}$$

for all $\xi \in \ell _2(X)$ , all $t\in \mathbb {R}$ , and all $a\in \mathrm {C}^{*}_u(X)$ .

For the last claim, notice that, as each $\sigma _t$ is the identity on $\ell _\infty (X)$ , the previous paragraph implies that each $u_t$ commutes with the elements of $\ell _\infty (X)$ . As $\ell _\infty (X)$ is a maximal abelian subalgebra of $\mathrm {C}^{*}_u(X)$ , this gives that $u_t\in \ell _\infty (X)$ for all $t\in \mathbb {R}$ .

Proof First notice that, as $t\in \mathbb {R}\mapsto \sigma _t(r_\xi )\in \mathrm {C}^{*}_u(X)$ is continuous, $t\in \mathbb {R}\mapsto u_t\xi \in \ell _2(X)$ is also continuous for all $\xi \in \ell _2(X)$ . Also, using (2.4), we have

$$\begin{align*}u_t(u_s\xi)=u_t(\sigma_s(r_\xi)\delta_x)=\sigma_t(\sigma_s(r_\xi))\delta_x=\sigma_{t+s}(r_\xi)\delta_x=u_{t+s}\xi\end{align*}$$

for all $\xi \in \ell _2(X)$ and all $t,s\in \mathbb {R}$ . Finally, as each $u_t$ is an element of $\ell _\infty (X)$ with norm at most one satisfying $u_{t}u_{-t}=1$ , this also shows that $u_t$ is a unitary.

By Claims 2.3 and 2.4, there is map $h\colon X\to \mathbb {R}$ such that

$$\begin{align*}u_t=e^{it\bar h}\ \text{ for all }\ t\in \mathbb{R}.\end{align*}$$

Therefore, by Claim 2.3, we have that $\sigma =\sigma _h$ . By Proposition 2.1, it follows that h must be coarse.

Proof of Proposition 1.4

(1) If h is bounded, $\bar h$ is a bounded operator on $\ell _2(X)$ . Therefore, the analyticity of $e^z$ gives that

$$\begin{align*}z\in \mathbb{C}\to e^{-iz\bar h}ae^{iz\bar h}\in \mathrm{C}^{*}_u(X)\end{align*}$$

is analytic for all $a\in \mathrm {C}^{*}_u(X)$ .

(2) Since $\mathrm {C}^{*}_u[X]$ is spanned by the subset of all $av_f$ , for $a\in \ell _\infty (X)$ and $f\colon A\subseteq X\to B\subseteq X$ a partial translation, it is enough to show that each such $av_f$ is analytic. Fix such a and $f\colon A\subseteq X\to B\subseteq X$ , and let $g\colon X\to \mathbb {R}$ be given by

$$\begin{align*}g(x)=\left\{\begin{array}{@{}ll} h(f(x))-h(x),& x\in A,\\ 0,& x\not\in A. \end{array}\right.\end{align*}$$

A simple computation gives that

$$\begin{align*}\sigma_{h,t}(av_f)=e^{it\bar g}av_f\end{align*}$$

for all $t\in \mathbb {R}$ . As $d(f(x),x)\leq r$ for all $x\in A$ , g is bounded. Then, the analyticity of $e^z$ implies the that

$$\begin{align*}z\in \mathbb{C}\to e^{iz\bar g}av_f\in \mathrm{C}^{*}_u(X)\end{align*}$$

is analytic; so, $av_f$ is analytic.

Proof of Theorem 1.5

As $\mathrm {C}^{*}_u[X]$ is dense in $\mathrm {C}^{*}_u(X)$ , it is enough to show that $\varphi (a)=\varphi (E(a))$ for all $a\in \mathrm {C}^{*}_u[X]$ . Moreover, as $\mathrm {C}^{*}_u[X]$ is the span of all $av_f$ , where $ a\in \ell _\infty (X)$ and f is a partial translation on X, it is enough to show that $\varphi (av_f)=0$ for all $a\in \ell _\infty (X)$ and all partial translations $f\colon A\subseteq X\to B\subseteq X$ such that $f(x)\neq x$ for all $x\in A$ ; fix a and f as such.

Let $r=\sup _{x\in A}d(x,f(x))$ ; as f is a partial translation, r is finite. As X is u.l.f., there is a partition

$$\begin{align*}A=A_1\sqcup\ldots\sqcup A_n,\end{align*}$$

such that each $A_i$ is $2r$ -separated, that is, $d(x,y)>2r$ for all $i\in \{1,\ldots , n\}$ and all distinct $x,y\in A_i$ . Therefore,

$$\begin{align*}d(x,f(y))\geq d(x,y)-d(y,f(y))>r\end{align*}$$

for all $i\in \{1,\ldots , n\}$ and all distinct $x,y\in A_i$ ; in particular, $x\neq f(y)$ . Moreover, as $f(x)\neq x$ for all $x\in A$ , this shows that

(2.5) $$ \begin{align} A_i\cap f(A_i)=\emptyset \end{align} $$

for all $i\in \{1,\ldots , n\}$ .

For each $i\in \{1,\ldots , n\}$ , let $f_i=f\restriction A_i$ . So, (2.5) implies that $\chi _{A_i}v_{f_i}=0$ for all $i\in \{1,\ldots , n\}$ . Therefore, since

$$\begin{align*}\chi_{A_i}\sigma_{h,i\beta}(av_{f_i})=\chi_{A_i}e^{-\beta \bar h}av_{f_i}e^{\beta\bar h}=e^{-\beta \bar h}a\chi_{A_i}v_{f_i}e^{\beta\bar h}=0,\end{align*}$$

we conclude that

$$\begin{align*}\varphi(av_{f_i})=\varphi( av_{f_i}\chi_{A_i})=\varphi(\chi_{A_i}\sigma_{h,i\beta}(av_{f_i}))=0.\end{align*}$$

Since $v_f=v_{f_1}+\ldots +v_{f_n}$ , this finishes the proof.

Proof Suppose first that $\varphi $ is a $(\sigma _{h},\beta )$ -KMS on $\mathrm {C}^{*}_u(X)$ . Let $f\colon A\to f(A)$ be a partial translation on X. Then, $\chi _{f(A)}=\chi _{f(A)}v_fv_f^{*}$ . As

$$\begin{align*}v_f^{*}\sigma_{h,i\beta}(\chi_{f(A)}v_f)=v_f^{*}e^{-\beta \bar h}\chi_{f(A)}v_fe^{\beta \bar h}=\chi_A e^{\beta (\overline{ h-h\circ f})},\end{align*}$$

the KMS condition gives that

$$\begin{align*}\varphi(\chi_{f(A)})=\varphi(v_f^{*}\sigma_{h,i\beta}(\chi_{f(A)}v_f))=\varphi\Big(\chi_A e^{\beta (\overline{ h-h\circ f})}\Big).\end{align*}$$

Suppose now that $\varphi $ satisfies (2.6). First, notice that as $\ell _\infty (X)$ is linearly generated by the characteristic functions on X, this implies that

(2.7) $$ \begin{align} \varphi(c ) =\varphi\Big(c_{\circ f} e^{\beta (\overline{ h-h\circ f})} \Big) \end{align} $$

for all partial translations f on X and all $c\in \ell _\infty (\mathrm {Im}(f))$ . By abuse of notation, we extend $\varphi $ to the whole $\mathrm {C}^{*}_u(X)$ and still denote it by $\varphi $ , that is, $\varphi =\varphi \circ E$ . In order to show that $\varphi \circ E$ is a $(\sigma _{h},\beta )$ -KMS state on $\mathrm {C}^{*}_u(X)$ , it is enough to show the KMS condition for elements of the form $av_f$ , where $a\in \ell _\infty (X)$ and f is a partial translation of X.

Fix $a,b\in \ell _\infty (X)$ and partial translations f and g on X. Let

$$\begin{align*}A=\Big\{x\in \mathrm{Dom}(f)\mid f(x)\in \mathrm{Dom}(g)\text{ and }g(f(x))=x\Big\}.\end{align*}$$

and notice that $g\restriction f(A)=(f\restriction A)^{-1}$ . We can then write

$$ \begin{align*} v_gav_f &= v_{g\restriction f(A)}av_{f\restriction A}+v_{g\restriction \mathrm{Dom}(g)\setminus f(A)}av_{f\restriction A}\\& \quad +v_{g\restriction f(A)}av_{f\restriction \mathrm{Dom}(f)\setminus A} +v_{g\restriction \mathrm{Dom}(g)\setminus f(A)}av_{f\restriction \mathrm{Dom}(f)\setminus A}\\&= v_{(f\restriction A)^{-1}}av_{f\restriction A}+v_{g\restriction f(A)}av_{f\restriction \mathrm{Dom}(f)\setminus A} +v_{g\restriction \mathrm{Dom}(g)\setminus f(A)}av_{f\restriction \mathrm{Dom}(f)\setminus A}. \end{align*} $$

Notice that the last two terms in the right handside of the equality above are in the kernel of the conditional expectation E. Therefore,

$$\begin{align*}E(bv_gav_f)=E(b v_{(f\restriction A)^{-1}}av_{f\restriction A}).\end{align*}$$

For this reason, it is enough to check the KMS condition for partial translations of X which are inverse of each other. For now on, assume that $g=f^{-1}$ .

Let us now show the KMS condition holds. Firstly, notice that

(2.8) $$ \begin{align} bv_fav_f^{*}=ba_{\circ f^{-1}}\text{ and } av_f^{*}e^{-\beta \bar h}bv_fe^{\beta \bar h}=a b_{\circ f} e^{\beta (\overline{ h-h\circ f})}. \end{align} $$

Then, letting $c=ba_{\circ f^{-1}}$ , we have that $c\in \ell _\infty (\mathrm {Im}(f))$ and

$$\begin{align*}c_{\circ f}=v_f^{*}cv_f=v_f^{*}bv_fav_f^{*}v_f=v_f^{*}bv_fa=ab_{\circ f.}\end{align*}$$

Therefore, (2.7) gives that

$$ \begin{align*} \varphi(bv_fav_f^{*})&=\varphi(ba_{\circ f^{-1}})\\ &=\varphi(c)\\ &=\varphi(c_{\circ f} e^{\beta (\overline{ h-h\circ f})} )\\ &=\varphi(a b_{\circ f} e^{\beta (\overline{ h-h\circ f})} )\\ &=\varphi\Big(av_f^{*}e^{-\beta \bar h}bv_fe^{\beta \bar h} \Big)\\ &=\varphi( av_f^{*}\sigma_{ h,i\beta}(bv_f) ). \end{align*} $$

This shows that $\varphi $ is a $(\sigma _{h},\beta )$ -KMS state on $\mathrm {C}^{*}_u(X)$ .

Proof of Theorem 2.7

We start recalling a well-known fact about uniform Roe algebras: a u.l.f. metric space has a positive unital trace if and only if it is amenable [Reference RoeRoe03, Theorem 4.6]. In fact, if $\mu $ is a nontrivial invariant mean on X, say $\mu (X)=1$ , and $E\colon \mathrm {C}^{*}_u(X)\to \ell _\infty (X)$ is the canonical conditional expectation, then

$$\begin{align*}\tau(a)=\int_XE(a) d\mu,\ \text{ for all }\ a\in \mathrm{C}^{*}_u(X),\end{align*}$$

defines a positive unital trace on $\mathrm {C}^{*}_u(X)$ . On the other hand, if $\tau $ is a positive unital trace on $\mathrm {C}^{*}_u(X)$ , then

$$\begin{align*}\mu(A)=\tau(\chi_A)\ \text{ for all }\ A\subseteq X\end{align*}$$

defines an invariant mean on X.

Suppose then that X is amenable and that $\tau $ is the trace on $\mathrm {C}^{*}_u(X)$ given by a nontrivial invariant mean $\mu $ on X as above. By Lemma 2.8, $\varphi _{\tau , e^{\beta \bar h}}$ satisfies the $(\sigma _{h},\beta )$ -KMS condition. Moreover, using the formula of $\tau $ , we have that

$$\begin{align*}\varphi_{\tau, e^{\beta \bar h}}(a)=\int_XE(a)e^{\beta \bar h} d\mu\ \text{ for all }\ a\in \mathrm{C}^{*}_u(X).\end{align*}$$

Therefore, $\varphi $ is positive and, as $t=\sup _{x\in X} |h(x)|<\infty $ , we have that

$$\begin{align*}\varphi_{\tau, e^{\beta \bar h}}(\chi_X)=\int_Xe^{\beta \bar h} d\mu\geq e^{-|\beta| t} \mu(X)>0.\end{align*}$$

Therefore, normalizing $\varphi $ , we obtain a $(\sigma _{h},\beta )$ -KMS state on $\mathrm {C}^{*}_u(X)$ .

Suppose now that $\varphi $ is a $(\sigma _{h},\beta )$ -KMS state on $\mathrm {C}^{*}_u(X)$ . By Lemma 2.8, $\tau _{\varphi ,e^{\beta \bar h}}$ satisfies the trace condition, that is, $\tau _{\varphi ,e^{\beta \bar h}}(ab)=\tau _{\varphi ,e^{\beta \bar h}}(ba)$ for all $a,b\in \mathrm {C}^{*}_u(X)$ . As $\varphi $ is positive and factors through the canonical conditional expectation $\mathrm {C}^{*}_u(X)\to \ell _\infty (X)$ (Theorem 1.5), $\tau _{\varphi , e^{\beta \bar h}}$ is also positive. Finally, it follows form our definition of t that

$$\begin{align*}\tau_{\varphi,e^{\beta\bar h}}(\chi_X)=\varphi(e^{-\beta\bar h})\geq \varphi(e^{-|\beta|t}\chi_{X})>0.\end{align*}$$

So, normalizing $\tau $ , we obtain a positive unital trace on X.

Proof Fix $x\in X$ . As h is unbounded, there is a sequence $(x_n)_n$ in X such that $\lim _n h(x_n)=\infty $ . Then, if $\varphi $ is a $(\sigma _{ h},\beta )$ -KMS state on $\mathrm {C}^{*}_u(X)$ , we have

$$\begin{align*}\varphi(e_{x,x})=\varphi(e_{x,x_n}e_{x_n,x})=\varphi(e_{x_n,x}\sigma_{h,i\beta}(e_{x,x_n}))=e^{\beta(h(x_n)-h(x))}\varphi(e_{x_n,x_n}).\end{align*}$$

As $(\varphi (e_{x_n,x_n}))_n$ is bounded and $\beta <0$ , we conclude that $\varphi (e_{x,x})=0$ by letting n go to infinity.

Proof of Theorem 1.6

Suppose $\varphi $ is a strongly continuous $(\sigma _{ h},\beta )$ -KMS state on $\mathrm {C}^{*}_u(X)$ . Fix $x_0\in X$ (this can be thought of as the “center” of X). Since all maps $(f_x\colon \{x_0\}\to \{x\})_{x\in X}$ are partial translations, the KMS condition gives us that

$$\begin{align*}\varphi(e_{x,x})=e^{-\beta(h(x)-h(x_0))}\varphi(e_{x_0,x_0})\end{align*}$$

for all $x\in X$ (see Theorem 2.5). As $\varphi $ is strongly continuous,

$$\begin{align*}1=\varphi(\chi_X)=\sum_{x\in X}\varphi(e_{x,x})=e^{\beta h(x_0)}\varphi(e_{x_0,x_0})\sum_{x\in X}e^{-\beta h(x)}. \end{align*}$$

So, $\varphi (e_{x_0,x_0})\neq 0$ and

$$\begin{align*}Z(\beta)=\sum_{x\in X}e^{-\beta h(x)}=\frac{1}{e^{\beta h(x_0)}\varphi(e_{x_0,x_0})}\end{align*}$$

must be finite (as well as independent on $x_0$ ). The formula for $\varphi $ in the statement of the theorem then follows immediately from the strong continuity of $\varphi $ .

Suppose now $Z(\beta )$ is finite and $\varphi $ is given as in the statement of the theorem. Clearly, $\varphi $ is a strongly continuous state on $\mathrm {C}^{*}_u(X)$ . Moreover, if $f\colon A\to B$ is a partial translation of X, then, by the formula of $\varphi $ , we have

$$ \begin{align*} \varphi(\chi_{f(A)})&=\frac{1}{Z(\beta)}\sum_{x\in f(A)}e^{-\beta h(x)}\\ &=\frac{1}{Z(\beta)}\sum_{x\in A}e^{-\beta h(f(x))}\\ & = \varphi(\chi_{A}e^{\beta(\overline{ h-h\circ f})}). \end{align*} $$

So, by Theorem 2.5, $\varphi $ is a $(\sigma _{h},\beta )$ -KMS state on $\mathrm {C}^{*}_u(X)$ .

Proof Let $f\colon A\subseteq X_0\to B\subseteq X_0$ be a partial translation. Then, there must be a partition $A=A_1\sqcup A_2$ such that $f(x)=x$ for all $x\in A_1$ and $|A_2|<\infty $ . As ${\varphi \restriction c_0(X_0)=0}$ , we have that

$$\begin{align*}\varphi(\chi_{f(A)})=\varphi(\chi_{f( A_1)}+\chi_{f( A_2)})=\varphi(\chi_{f( A_1)})=\varphi(\chi_{ A_1}).\end{align*}$$

Similarly, we have

$$\begin{align*}\varphi\Big(\chi_Ae^{\beta(\overline{ h-h\circ f})}\Big)=\varphi\Big(\chi_{ A_1} e^{\beta(\overline{ h-h\circ f})}\Big)=\varphi(\chi_{ A_1}).\end{align*}$$

The result then follows from Theorem 2.5.

Proof This is a straightforward consequence of Theorem 1.6.

Proof This follows immediately from Propositions 2.12 and Corollary 2.14.

Proof If $(X,d)$ is infinite and u.l.f., then X is unbounded. Hence, we can inductively pick a sequence $(x_i)_{i\in \mathbb {N}}$ in X such that

$$\begin{align*}d(x_k,x_\ell)\geq \max _{i,j<\ell}d(x_i,x_j)+\ell\end{align*}$$

for all $\ell>k$ in $\mathbb {N}$ . The set $A=\{x_i\mid i\in \mathbb {N}\}$ is clearly thin.Footnote ⁸

Proof By Lemma 4.7, it is enough to show that $(g^{-1}\circ f)[A]\cap B$ is finite. As the composition of partial translations is still a partial translation, it is enough to show that $ f[A]\cap B$ is finite for any partial translation f of X. Fix such f and, replacing A with $A\cap \mathrm {Dom}(f)$ , we also assume that $A\subseteq \mathrm {Dom}(f)$ . Let us show that $f(A)\cap B$ is finite. Set

$$\begin{align*}A_0=\{x\in A\mid f(x)=x\} \ \text{ and }\ A_1=A\setminus A_0.\end{align*}$$

Then, as $A\cap B=\emptyset $ , we have

$$\begin{align*}f(A)\cap B=f(A_0\cup A_1)\cap B=f(A_1)\cap B.\end{align*}$$

Let $f_1=f\restriction A_1$ . Then $f_1$ has no fixed points and

$$\begin{align*}f_1(A)\cap B\subseteq f_1[C]\cap C.\end{align*}$$

Since C is thin, $f_1[C]\cap C$ must be finite. So, $f(A)\cap B$ is finite.

Proof By Proposition 4.3, it is enough to notice that $\chi _L$ is in the center of $\mathrm {Q}^{*}_u(X)$ . Hence, since $\mathrm {C}^{*}_u[X]$ is dense in $\mathrm {C}^{*}_u(X)$ and spanned by $av_f$ , where $a\in \ell _\infty (X)$ and f is a partial translation of X, we only need to show that $\chi _{L}$ commutes with $w_f=\pi (v_f)$ for all partial translations f of X. Fix such partial translation f and let $A=\mathrm {Dom}(f)$ and $B=\mathrm {Im}(f)$ . Then, $w_f= \chi _{\hat B} w_f\chi _{\hat A}$ and

(4.1) $$ \begin{align} w_f\chi_L= w_f\chi_{\hat A\cap L}=\chi_{\hat B\cap f[L]}w_f =\chi_{f[L]}w_f \end{align} $$

notice that $f[L]=\hat B\cap L$ . Indeed, since L is invariant and f is a partial translation, $f[L]\subset \hat B\cap L$ . On the other hand, as $f^{-1}$ is also a partial translation, we have $f^{-1}[L]\subseteq L$ . Hence, as $\hat B\cap L\subseteq f[f^{-1}[L] ]$ , we also have $ \hat B\cap L\subseteq f[L]$ . We can then conclude from (4.1) that $w_f\chi _L=\chi _Lw_f$ . As the partial translation f was arbitrary, we conclude that $\chi _L\in C(\nu X)$ as desired.

Proof Since $\omega ,\omega '\in \bar C$ , it follows that $C\in \omega $ and $C\in \omega '$ . As $\omega \neq \omega '$ , there is $D\subseteq X$ such that $D\in \omega $ and $D\not \in \omega '$ . Hence,

$$\begin{align*}A=C\cap D\in \omega\ \text{ and }\ B=C\setminus D\in \omega'.\end{align*}$$

Therefore, $\omega \in \hat A$ and $\omega '\in \hat B$ . Let $\mathcal {PT}$ denote the set of all partial translations of X and define

$$\begin{align*}U=\bigcup_{f\in \mathcal{PT}}\widehat{f[A]}\ \text{ and }\ V=\bigcup_{f\in \mathcal{PT}}\widehat{f[B]}.\end{align*}$$

Clearly, U and V are open, invariant and contain $\omega $ and $\omega '$ , respectively. We only need to notice they are also disjoint. For that, notice that Proposition 4.10 implies that $f[A]\cap g[B]$ is finite for all $f,g\in \mathcal {PT}$ . But then $\widehat {f[A]}\cap \widehat {g[B]}=\emptyset $ for all $f,g\in \mathcal {PT}$ , which in turn implies that $U\cap V=\emptyset $ .

Proof Let $p\colon \hat X\to \nu X$ be the continuous surjection such that the canonical identification of $C(\nu X)$ with a $\mathrm {C}^{*}$ -subalgebra of $C(\hat X)$ is given by the map

$$\begin{align*}a\in C(\nu X)\mapsto a\circ p\in C(\hat X ).\end{align*}$$

Let $C\subseteq X$ be an infinite thin subset given by Lemma 4.9. As $\hat C$ is the set of all nonprincipal ultrafilters on C and C is countable, we have that $|\hat C|=2^{2^{\aleph _0}}$ . Therefore, in order to obtain that $\nu X$ has $2^{2^{\aleph _0}}$ elements, it is enough to show that p is injective on $\hat C$ .

Let $\omega ,\omega '\in \hat C$ be distinct. By Lemma 4.13, there are disjoint invariant open subsets $U,V\subseteq \hat X$ containing $\omega $ and $\omega '$ , respectively. As $\beta X$ is extremely disconnected, $\bar U$ is clopen in $\hat X$ which implies that the characteristic function of $\bar U$ , $\chi _{\bar U}$ , is a continuous function in $C(\hat X)$ . As $\bar U$ is invariant, Lemma 4.12 shows that $\chi _{\bar U}\in C(\nu X)$ . Therefore, since we clearly have $\chi _{\bar U}(\omega )=1$ and $\chi _{\bar U}(\omega ')=0$ , this shows that $p(\omega )\neq p(\omega ')$ .

Proof The statement for the uniform Roe corona follows from Theorems 4.4 and 4.14. The statement for the uniform Roe algebra is then a consequence of Proposition 1.8.

Assumption 5.2 Let $n\in \mathbb {N}$ and let $T_n$ be the n-branching tree endowed with the shortest path metric, denoted by d. Let $h\colon T_n\to \mathbb {R}$ be the function given by $h(x)=d(x,\emptyset )$ for all $x\in T_n$

Proof For each j, denote by $b_j=\varphi (\chi _{\{j\}})$ . Then, as $\varphi \restriction c_0$ has norm 1,

$$\begin{align*}\sum_{j=1}^\infty b_j= \lim_k\sum_{j=1}^kb_j=\lim_k \varphi(\chi_{\{1,\dots,k\}})=1. \end{align*}$$

Let a be a positive element in $\ell _\infty $ with norm at most 1. Then, as $\varphi $ is positive, we have that

$$\begin{align*}\sum_{j=1}^k a_jb_j = \varphi(a \chi_{\{1,\dots,k\}}) \leq \varphi( a)\ \text{ for all }\ k\in\mathbb{N}. \end{align*}$$

So, upon taking the limit as $k\to \infty $ , we get

$$\begin{align*}\sum_{j=1}^\infty a_jb_j\leq \varphi( a). \end{align*}$$

Applying the same reasoning to $1-a$ we deduce that

$$\begin{align*}\varphi(a) = 1-\varphi(1-a) \leq 1 - \sum_{j=1}^\infty (1-a_j)b_j = \sum_{j=1}^\infty a_jb_j. \end{align*}$$

Hence,

$$\begin{align*}\varphi( a)= \sum_{j=1}^\infty a_jb_j.\end{align*}$$

Now, for an arbitrary $a\in \ell _\infty $ , splitting a into its real and imaginary parts, and splitting each such parts into their positive and negative parts, the previous paragraph imply that $\varphi ( a)= \sum _{j=1}^\infty a_jb_j$ , so the lemma follows.

Proof Suppose $\varphi $ is a $(\sigma _{h},\beta )$ -KMS state on $\mathrm {C}^{*}_u(T_n)$ . Notice that, for each $y\in T_n$ , the map $f\colon T_n\to T_n$ given by $f(x)=x^{\smallfrown }y$ , for all $x\in T_n$ , is a partial translation; indeed, $d(x,f(x))=|y|$ for all $x\in T_n$ . So, each $v_f$ belongs to $\mathrm {C}^{*}_u[T_n]$ . Then, for each $y\in T_n$ , we have

$$ \begin{align*} \sigma_{h,i\beta}(v_f^{*})&=e^{-\beta\bar h}v_f^{*}e^{\beta\bar h}\\ &=e^{-\beta\bar h}\Big(\mathrm{SOT}\text{-}\sum_{x\in X}e_{x,x^\smallfrown y}\Big)e^{\beta\bar h}\\ &=e^{\beta|y|}v_f^{*}. \end{align*} $$

For each $y\in T_n$ , set

$$\begin{align*}T_n^\smallfrown y=\{x\in T_n\mid x=z^\smallfrown y\text{ for some }z\in T_n\}.\end{align*}$$

Hence, as $\chi _{T_n}=v_f^{*}v_f$ and $\chi _{T_n^\smallfrown y}=v_fv_f^{*}$ , we must have

$$\begin{align*}1=\varphi(\chi_{T_n})=\varphi(v_f^{*}v_f)= \varphi(v_f\sigma_{h,i\beta}(v_f^{*}))=e^{\beta |y|}\varphi(\chi_{T_n^\smallfrown y})\end{align*}$$

for all $y\in T_n$ ; which implies

(5.1) $$ \begin{align}\varphi(\chi_{T_n^\smallfrown y})=e^{-\beta |y|} \ \text{ for all } \ y\in T_n. \end{align} $$

Since for each $y\in T_n$ , we have

$$\begin{align*}\{y\}=T_n^\smallfrown y\setminus \bigsqcup_{k=1}^n T_n^\smallfrown k^\smallfrown y,\end{align*}$$

where “ $\bigsqcup $ ” denotes disjoint union, (5.1) implies that

(5.2) $$ \begin{align} \varphi(e_{y,y})=\varphi\Big(\chi_{T_n^\smallfrown y}-\sum_{k=1}^n\chi_{T_n^\smallfrown k^\smallfrown y}\Big)=e^{-\beta|y|}-ne^{-\beta (|y|+1)} \end{align} $$

for all $y\in T_n$ .

As $\varphi $ is positive, each $\varphi (e_{y,y})$ must be positive. So, (5.2) gives that

$$\begin{align*}e^{-\beta|y|}\geq ne^{-\beta (|y|+1)}\ \text{ for all }\ y\in T_n.\end{align*}$$

Solving for $\beta $ , this implies $\beta \geq \log (n)$ . Moreover, as (5.2) must hold regardless of $\beta $ , this also shows that the $(\sigma _{h},\log (n))$ -KMS states on $\mathrm {C}^{*}_u(T_n)$ all vanish on $c_0(T_n)$ . Since such states factors through $\ell _\infty (T_n)$ (Theorem 1.5), (2) follows.

We must now show that if $\beta \geq \log (n)$ , then $(\sigma _{h},\beta )$ -KMS states exist. This will however be an immediate consequence of (1). Indeed, the set of all $\beta $ ’s for with $(\sigma _{h},\beta )$ -KMS states exist is always a closed set (see [Reference Bratteli and RobinsonBR97, Proposition 5.3.25]).

We now show (1) holds. For this, suppose $\beta>\log (n)$ and let us show that any given $(\sigma _h,\beta )$ -KMS state $\varphi $ must have the required form. Notice that $\varphi \restriction \ell _\infty (T_n)$ is a state on $\ell _\infty (T_n)$ . Moreover, the computations above show that

(5.3) $$ \begin{align} \varphi (a)=\sum_{y\in T_n} a_{y}\Big(e^{-\beta|y|}-ne^{-\beta (|y|+1)}\Big) \end{align} $$

for all $a=(a_y)_y\in c_0(T_n)$ . Hence, an easy computation gives

$$\begin{align*}\lim_{F,\mathcal{F}}\varphi(\chi_F)=1,\end{align*}$$

where $\mathcal {F}$ is the net of all finite subsets of $T_n$ ordered by reverse inclusion. Therefore, it follows that $\|\varphi \restriction c_0(T_n)\|=1$ and, by Lemma 5.3, $\varphi \restriction \ell _\infty $ is strongly continuous. This implies that (5.3) holds for all $a=(a_y)_y\in \ell _\infty (T_n)$ . In order to notice that this holds for arbitrary elements of $\mathrm {C}^{*}_u(T_n)$ , let $E\colon \mathrm {C}^{*}_u(X)\to \ell _\infty (X)$ be the canonical conditional expectation and recall that, by Theorem 1.5, we have $\varphi =\varphi \circ E$ . This proves the uniqueness part of (1).

We are left to notice that a $\varphi $ given by the formula above is indeed a $(\sigma _{h},\beta )$ -KMS state on $\mathrm {C}^{*}_u(T_n)$ . This will be done by using Theorem 2.5.Footnote ¹⁰ So, let $f\colon A\to f(A)$ be a partial translation on X. On one hand, we have that