Convention 3.1.1. In order to avoid any confusion stemming from the abuse of notation of denoting by the same symbol — h usually — both a map $ X\to \mathbb {R} $ and its interpretation as an unbounded operator on $\ell _2(X)$ , we write $h=(h_x)_{x\in X}$ when h is being seen as a function $h\colon x\in X\mapsto h_x\in \mathbb {R} $ . If h is being interpreted as an operator on $\ell _2(X)$ , its evaluation at a vector $\xi \in \ell _2(X)$ is written as $h\xi $ . Also, $\operatorname {\mathrm {dom}} (h)$ will always refer to the domain of h as an operator on $\ell _2(X)$ (the domain of h as a function $X\to \mathbb {R} $ is simply X and this will not be a point of relevance in these notes).

Proof. It is evident that $\operatorname {\mathrm {dom}} (h)$ contains $c_{00}(X)$ , and we leave it for the reader to prove the standard fact that h is a closed operator. On the other hand, given $\xi \in \operatorname {\mathrm {dom}} (h)$ , we have that

$$\begin{align*}h\xi =(h_x\xi _x)_{x\in X} = \sum _{x\in X}h_x\xi _x\delta _x = \sum _{x\in X}\langle \xi , \delta _x\rangle h\delta _x, \end{align*}$$

so h is admissible. In order to show that $\operatorname {\mathrm {dom}} (h)$ is invariant under partial translations, suppose that $h\colon X\to \mathbb {R} $ is coarse and that $\xi =(\xi _x)_{x\in X}$ is in the domain of h, seen as an operator on $\ell _2(X)$ . So,

$$\begin{align*}\| h\xi \|^2=\sum _{x\in X}|h_x\xi _x|^2<\infty. \end{align*}$$

Let f be a partial translation of X. As h is coarse,

$$\begin{align*}s := \sup _{x\in \operatorname{\mathrm{dom}} (f)}|h_{f(x)}-h_x|<\infty. \end{align*}$$

Then, by Minkowski’s inequality, we have

$$ \begin{align*} \left (\sum _{x\in \operatorname{\mathrm{dom}} (f) }|h_{f(x)}\xi _x|^2\right )^{1/2}\leq & \left (\sum _{x\in \operatorname{\mathrm{dom}} (f) }|\xi _x|^2\cdot |h_{f(x)}-h_{x}|^2\right )^{1/2}\\ &\quad + \left (\sum _{x\in \operatorname{\mathrm{dom}} (f) }|h_x\xi _x|^2\right )^{1/2}\\ &\leq s\|\xi \|+\| h\xi \|\\ & < \infty. \end{align*} $$

This shows that $v_f\xi $ is in the domain of h and we are done.

Proof. (1) $\Rightarrow $ (2): Assume $h\colon X\to \mathbb {R} $ is coarse. By Proposition 3.1.4, $\operatorname {\mathrm {dom}} (h)$ is invariant under partial translations. Fix $r>0$ . As $h\colon X\to \mathbb {R} $ is coarse, there is $s>0$ such that $|h_x-h_y|\leq s$ for all $x,y\in X$ with $d(x,y)\leq r$ . Hence, if f is a partial r-translation of X, then $|h_{f(x)}-h_x|\leq s$ for all $x\in \operatorname {\mathrm {dom}} (f)$ . Since $[h,v_f](\delta _x)=(h_{f(x)}-h_x)\delta _{f(x)}$ for all $x\in \operatorname {\mathrm {dom}} (f)$ , it follows that

$$ \begin{align*} \| [ h, v_f] \| \leq \sup _{x\in \operatorname{\mathrm{dom}} (f)}|h_{f(x)}-h_x|\leq s. \end{align*} $$

(2) $\Rightarrow $ (3): This is immediate.

(3) $\Rightarrow $ (1): Suppose $h\colon X\to \mathbb {R} $ is not coarse. So, there is $r>0$ and sequences $(x_n)_n$ and $(y_n)_n$ in X such that $d(x_n,y_n)\leq r$ for all $n\in \mathbb {N} $ and

(3.1.6) $$ \begin{align} \lim _n|h_{x_n}-h_{y_n}|=\infty. \end{align} $$

As X is u.l.f., going to subsequences if necessary, we can assume that $(x_n)_n$ and $(y_n)_n$ are both sequences of distinct elements of X. Therefore, letting

$$\begin{align*}A=\{x_n\ :\ n\in \mathbb{N} \}\ \text { and }\ B=\{y_n\ :\ n\in \mathbb{N} \}, \end{align*}$$

the map $f\colon A\to B$ given by $f(x_n)=y_n$ for all $n\in \mathbb {N} $ defines a partial translation, a partial r-translation to be precise. However, we have

$$\begin{align*}\|[h,v_f] \|\geq \|[ h,v_f]\delta _{x_n}\|=|h_{f(x_n)}-h_{x_n}| \end{align*}$$

for all $n\in \mathbb {N} $ . By (3.1.6), the latter goes to infinity. But this contradicts the hypothesis.

Proof. As $\sigma _h$ is a flow, h must be coarse (Proposition 1.1.2). Let us first show that the first statement indeed implies the second. Since $\mathrm {C}^*_u [X]$ is the linear span of $av_f$ , for $a\in \ell _\infty (X)$ , and f a partial translation of X, it is enough to notice that the map (3.1.8) is differentiable at zero for elements of the form $av_f$ . Finally, as $\sigma _{h,t}(av_f)=a\sigma _{h,t}(v_f)$ , we see that the second statement indeed follows from the first one.

Fix now a partial translation f of X. An elementary calculation shows that

(3.1.9) $$ \begin{align} \sigma _{h,t}(v_f)\delta _x = e^{it (h(f(x))-h(x))}v_f\delta _x, \end{align} $$

if x is in $\operatorname {\mathrm {dom}} (f)$ , while $\sigma _{h,t}(v_f)\delta _x=0$ otherwise. Denoting by $h\circ f-h$ the function on X given by

$$\begin{align*}(h\circ f-h)(x) = \left \{ \begin{matrix} h(f(x))-h(x), &\ \ \ \ \text {if } x\in \operatorname{\mathrm{dom}} (f), \cr 0, &\text {otherwise, }\hfill \end{matrix}\right. \end{align*}$$

it follows from (3.1.9) that $\sigma _{h,t}(v_f) = v_fe^{it(h\circ f-h)}$ . Observing that $h\circ f-h$ is a bounded function, thanks to h being coarse, we deduce that $\sigma _{h,t}(v_f)$ is differentiable as a function of the variable t, and that its derivative at zero coincides with $iv_f(h\circ f-h)$ . Another elementary calculation shows that

$$\begin{align*}v_f(h\circ f-h)= [h, v_f], \end{align*}$$

from where the conclusion follows.

Proof. The “if” part being immediate, we move on to the “only if” direction. Assuming that h is a coarse operator we therefore suppose, by way of contradiction, that the supremum in the statement is infinite.

Proof. To see this, we fix $M>0$ , and we choose a partial r-translation g (which may or may not have finite domain), such that $\| [h,v_g]\|>M$ . Since $[h,v_g]$ is bounded on $\operatorname {\mathrm {dom}} (h)$ , by hypothesis, and since $c_{00}(X)$ is dense in $\operatorname {\mathrm {dom}} (h)$ , we may pick a unit vector $\xi $ in $c_{00}(X)$ , such that

(4.1.6) $$ \begin{align} M < \| [h,v_g]\xi \| = \| hv_g\xi - v_gh\xi \|. \end{align} $$

Fixing $\varepsilon>0$ , let C be a finite subset of X such that

$$\begin{align*}\xi =p_C\xi ,\quad \text {and}\quad \| h\xi -p_Ch\xi \| <\varepsilon. \end{align*}$$

Letting f be the restriction of g to $\operatorname {\mathrm {dom}} (g)\cap C$ , we then have that $v_{f} = v_gp_C$ . Moreover

$$\begin{align*}v_g\xi = v_{f}\xi ,\quad \text {and}\quad \| v_gh\xi - v_{f}h\xi \| <\varepsilon. \end{align*}$$

So, upon substituting g for f, the right-hand side of (4.1.6) changes by at most $\varepsilon $ , so the inequality is preserved as long as we choose $\varepsilon $ sufficiently small.

Slightly changing tack, we now make the following:

Proof. Indeed, given such an r-partial translation f, we have that

$$ \begin{align*} \| [h,v_f]\| & = \| hv_f-v_fh \| \\ & \leq \| hv_f \| + \| v_fh \| \\ & \leq \| hp_{B_r(C)} \|+ \| p_{C}h \|, \end{align*} $$

the last inequality due to the fact that $\operatorname {\mathrm {dom}} (f)\subseteq C$ , and hence that $\operatorname {\mathrm {ran}} (f)\subseteq B_{r}(C)$ , the ball of radius r around C, because r-partial translations are not allowed to move points more than r apart. As h is necessarily bounded on the finite-dimensional space $\ell _2(B_r(C))$ , $\| hp_{B_r(C)} \|$ is finite. In order to estimate $\| p_Ch \|$ , notice that

$$ \begin{align*}\| p_Ch\|\leq \|p_Chp_C\|+\|p_Chp_{X\setminus C}\|\leq \|hp_{C}\|+ \|p_Chp_{X\setminus C}\|.\end{align*} $$

As before, $\|hp_C\|$ is finite and, since $p_Chp_{X\setminus C}=[h,p_{X\setminus C}]p_{X\setminus C}$ , the fact that h is coarse implies that $\|p_Chp_{X\setminus C}\|$ is also finite. This completes the proof of the claim.

Our next claim is a strengthening of Claim 4.1.7.

Proof. Given $M>0$ , let N be the supremum referred to in Claim 4.1.7 relative to the finite set $B_r(C)$ , and, using Claim 4.1.5, choose some r-partial translation g, with finite domain, and such that $\| [h,v_g]\|>M+N$ . Letting $g_1$ and $g_2$ denote the restrictions of g to $\operatorname {\mathrm {dom}} (g)\cap B_r(C)$ , and $\operatorname {\mathrm {dom}} (g)\setminus B_r(C)$ , respectively, we see that $v_g=v_{g_1}+v_{g_2}$ , so

$$\begin{align*}\| [h,v_{g_2}]\| = \| [h,v_g-v_{g_1}]\| \geq \| [h,v_g]\| - \| [h,v_{g_1}]\|> M + N - N = M, \end{align*}$$

Recalling that $\operatorname {\mathrm {dom}} (g_2)\subseteq X\setminus B_r(C)$ , we have that $\operatorname {\mathrm {ran}} (g_2)\subseteq X\setminus C$ , for the same reason observed above that r-partial translations are not allowed to move points more than r apart.

From now on we fix a sequence $(f_n)_n$ as above.

Claim 4.1.10. For every finite subset $C\subseteq X$ , one has that

$$\begin{align*}\lim _{n\to \infty }\| p_{X\setminus C}[h, v_{f_n}]p_{X\setminus C} \| =\infty. \end{align*}$$

Proof. In order to verify this claim, observe that $\ell _2(C)$ is a finite dimensional space, hence $p_Ch$ and $hp_C$ are bounded operators on the domain of h, say with

$$\begin{align*}\| p_Ch\| , \ \| hp_C\| \leq M_C. \end{align*}$$

Therefore, for every n, we have that

$$\begin{align*}\| [h, v_{f_n}]p_C\| = \| h v_{f_n}p_C\| + \| v_{f_n}hp_C\| \leq M_{B_r(C)} + M_C, \end{align*}$$

and similarly $\| p_C[h, v_{f_n}]\| \leq M_{B_r(C)} + M_C$ . Therefore

$$ \begin{align*} \| p_{X\setminus C}[h, v_{f_n}] p_{X\setminus C}\| & = \| (1-p_C)[h, v_{f_n}](1-p_C)\| \\ & \geq \| [h, v_{f_n}]\| - \| [h, v_{f_n}]p_C\| \\ &\quad - \| p_C[h, v_{f_n}]\| - \| p_C[h, v_{f_n}]p_C\| \\ &\geq \| [h, v_{f_n}]\| - 3(M_{B_r(C)} + M_C), \end{align*} $$

from where the claim follows.

Claim 4.1.11. Upon passing to a subsequence of $(f_n)_n$ , there is a sequence $(X_n)_n$ of finite and pairwise disjoint subsets of X, such that both $\operatorname {\mathrm {dom}} (f_n)$ and $\operatorname {\mathrm {ran}} (f_n)$ are contained in $X_n$ , and

$$\begin{align*}\lim _{n\to \infty }\| p_{X_n}[h, v_{f_n}]p_{X_n}\| =\infty. \end{align*}$$

In order to conclude the proof, we let $(f_n)_n$ be as in Claim 4.1.10, and we consider the r-partial translation defined on $\bigsqcup _{n\in {\mathbb N}}\operatorname {\mathrm {dom}} (f_n)$ to be a common extension of all of the $f_n$ .

Observing that, for $n\neq m$ , one has that $\operatorname {\mathrm {dom}} (f_m)\subseteq X_m$ , and that $X_m$ and $X_n$ are disjoint, it follows that $v_{f_m}p_{X_n}=0$ , and similarly that $p_{X_n}v_{f_m}=0$ , so

$$\begin{align*}p_{X_n}[h, v_{f}]p_{X_n} = p_{X_n}[h, v_{f_n}]p_{X_n}, \end{align*}$$

whence, for every n,

$$\begin{align*}\| [h, v_{f}]\| \geq \| p_{X_n}[h, v_{f}]p_{X_n}\| = \| p_{X_n}[h, v_{f_n}]p_{X_n}\|. \end{align*}$$

Observing that the right hand side goes to $\infty $ , as $n\to \infty $ , by Claim 4.1.11, we get a contradiction with the assumption that $[h, v_{f}]$ is a bounded operator.

Proof. (1) Suppose $E(h)$ is not coarse. Then, there are $r>0$ , and sequences $(x_n)_n$ and $(y_n)_n$ in X such that

(4.1.14) $$ \begin{align} \lim _n|E(h)_{x_n}-E(h)_{y_n}|=\infty \end{align} $$

and $d(x_n,y_n)\leq r$ for all $n\in \mathbb {N} $ . As X is u.l.f., going to subsequences if necessary, we can assume that $(x_n)_n$ and $(y_n)_n$ are sequences of distinct elements. Let $A=\{x_n\ :\ n\in \mathbb {N} \}$ , $B=\{y_n\ :\ n\in \mathbb {N} \}$ , and $f\colon A\to B$ be the partial translation given by $f(x_n)=y_n$ for all $n\in \mathbb {N} $ . Then,

$$ \begin{align*} \left \|[h,v_f] \right \|&\geq \left \|[h,v_f]\delta _{x_n}\right \|\\ & = \left \|\sum _{z\in X} \langle h\delta _{y_n},\delta _z\rangle \delta _z - \langle v_fh\delta _{x_n},\delta _z\rangle \delta _z\right \| \\ &\geq | \langle h\delta _{y_n},\delta _{y_n}\rangle - \langle h\delta _{x_n},\delta _{x_n}\rangle |\\ &=|E(h)_{y_n}-E(h)_{x_n}| \end{align*} $$

for all $n\in \mathbb {N} $ . By (4.1.14), this shows that $ [h,v_f]$ is unbounded on $\operatorname {\mathrm {dom}} (h)$ , which contradicts the fact that h is a coarse operator.

(2) As h is a coarse operator, Proposition 4.1.4 gives $s>0$ , such that

$$\begin{align*}\|[h,p_A] \|\leq s \end{align*}$$

for all $A\subseteq X$ . Since $p_{X\setminus A}hp_A = [h, p_A]p_A$ , the result follows.

Proof. As h is a coarse operator, Proposition 4.1.13(2) implies that

$$\begin{align*}M=\sup \{\|p_{X\setminus A}hp_A\|\ : \ A\subseteq X\}\end{align*}$$

is finite. Suppose towards a contradiction that the theorem fails and let $\xi \in \operatorname {\mathrm {dom}} (h)$ be an offender. As h is a closed operator, the series in the statement of the theorem not only fails to coincide with $h\xi $ , but it does not even converge. Consequently that series also fails to satisfy Cauchy’s criterion for the summability of a series, whence, letting $\xi _F=\sum _{x\in F}\langle \xi ,\delta _x\rangle \delta _x$ for each $F\subseteq X$ , we can find $\varepsilon>0$ , and a sequence $(F(n))_n$ of finite disjoint subsets of X such that

(4.1.16) $$ \begin{align} \|h\xi _{F(n)}\|\geq \varepsilon \ \text { for all }\ n\in \mathbb{N}. \end{align} $$

As $\lim _n\xi _{F(n)}=0$ , going to a subsequence if necessary, we can assume that $\zeta =\sum _{n\in \mathbb {N} }\xi _{F(n)}$ has norm at most $\varepsilon /(3M)$ . In particular, (4.1.16) can be improved as follows:

$$ \begin{align*} \|p_{F(n)}h\xi _{F(n)}\|\geq \|h\xi _{F(n)}\|-\|p_{X\setminus F(m)}hp_{F(n)}\xi _{F(n)}\|\geq \frac {2\varepsilon }{3} \end{align*} $$

for all $n\in \mathbb {N} $ . Since h is a coarse operator, its domain is invariant under partial translations and, in particular, invariant under the projections $p_A$ , $A\subseteq X$ . So, $\zeta \in \operatorname {\mathrm {dom}} (h)$ , since $\zeta $ is the image of $\xi $ under such projection. However,

$$ \begin{align*} \|p_{F(n)}h\zeta \| & \geq \|p_{F(n)}h\xi _{F(n)}\|-\left \|p_{F(n)}hp_{X\setminus F(n)}\left (\sum _{m\in \mathbb{N} \setminus \{n\}}\xi _{F(m)}\right )\right \|\\ &\geq \frac {\varepsilon }{3} \end{align*} $$

for all $n\in \mathbb {N} $ . As $(F(n))_n$ are disjoint sets, this shows that $h\zeta $ has infinite norm, a contradiction.

Proof. This is an immediate consequence of Proposition 2.1.2 and Theorem 4.1.15.

Proof. Suppose there is an operator $h'$ on $\ell _2(X)$ with finite propagation so that ${h-h'}$ is bounded. In particular, $E(h-h')=E(h)-E(h')$ is also bounded. Recall that, as X is u.l.f., an operator w on $\ell _2(X)$ with finite propagation is bounded if and only if $\sup _{x,y\in X}|\langle w\delta _x,\delta _y\rangle |<\infty $ (Proposition 2.3.1). By Proposition 4.1.13(2) above,

$$\begin{align*}\sup \left \{|\langle h\delta _x,\delta _y\rangle |\ :\ x,y\in X\text { with } x\neq y\right \}<\infty , \end{align*}$$

so, since $h-h'$ is bounded, we have

$$\begin{align*}\sup \left \{|\langle h'\delta _x,\delta _y\rangle |\ :\ x,y\in X\text { with } x\neq y\right \}<\infty. \end{align*}$$

Hence,

$$\begin{align*}\sup _{x,y\in X}|\langle (h'-E(h'))\delta _x,\delta _y\rangle |<\infty. \end{align*}$$

As $h'$ has finite propagation, so has $h'-E(h')$ . So, $h'-E(h')$ is bounded. Finally, as the equality

$$\begin{align*}h-E(h)=h-h'+h'-E(h')+E(h')-E(h) \end{align*}$$

holds on $c_{00}(X)$ , it follows that $h-E(h)$ is bounded on $c_{00}(X)$ . Therefore, since both h and $E(h)$ are coarse operators (for $E(h)$ , this is a consequence of Propositions 3.1.5 and 4.1.13(2)), Theorem 4.1.15 implies that $h-E(h)$ is bounded on all of its domain.

Proof of Proposition 1.3.7

This proposition is simply an interpretation of [Reference Bratteli and RobinsonBR87, Proposition 3.2.55] to our setting. Precisely, suppose $\sigma $ is a coarse flow, i.e., $\sigma $ is a flow and

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

is differentiable at zero for all partial translations f on X.Footnote ¹³ Since flows on $\mathrm {C}^*_u (X)$ are implemented by self-adjoint operators, we must have $\sigma =\sigma _h$ for some self-adjoint h on $\ell _2(X)$ (Proposition 1.3.3). The implication (1) $\Rightarrow $ (3) of [Reference Bratteli and RobinsonBR87, Proposition 3.2.55] then says that $\operatorname {\mathrm {dom}} (h)$ is invariant under partial translations and that $[h,v_f]$ is bounded on $\operatorname {\mathrm {dom}} (h)$ , i.e., h is a coarse operator. Regarding our usage of [Reference Bratteli and RobinsonBR87, Proposition 3.2.55] here, we point out that this result says that there is a core of h which is invariant by it and in which the restriction of $[h,v_f]$ to it is bounded. But, as noticed in the first paragraph of the proof of [Reference Bratteli and RobinsonBR87, Proposition 3.2.55], closure arguments imply that one can take the core to be the whole $\mathrm {dom}(h)$ .

Conversely, suppose that h is a coarse operator and fix a partial translation f of X. Then $\operatorname {\mathrm {dom}} (h)$ is f-invariant and $[h,v_f]$ is bounded on $\operatorname {\mathrm {dom}} (h)$ . We are then in the setting of item (3) of [Reference Bratteli and RobinsonBR87, Proposition 3.2.55] and the implication (3) $\Rightarrow $ (1) therein says that

$$\begin{align*}t\in \mathbb{R} \mapsto \sigma _{h,t}(v_f)\in \mathrm{C}^*_u(X) \end{align*}$$

is differentiable, and hence continuous at zero. As the $v_f$ span a dense subset of $\mathrm {C}^*_u (X)$ , by definition, we deduce that $\sigma _h$ is a flow, which is coarse by what was said just above.

Proof. As in the proof of Proposition 1.3.7, this is given by [Reference Bratteli and RobinsonBR87, Proposition 3.2.55].

Proof. Fix $x_0\in X$ and let $h(x)=d(x,x_0)$ for all $x\in X$ . So, $h\colon X\to \mathbb {R} $ is a coarse map and $\sigma _h$ is a diagonal flow on $\mathrm {C}^*_u (X$ ) (Proposition 1.1.2). Clearly, $ \sigma _{uhu*}$ is also a flow on $\mathrm {C}^*_u (X)$ for all unitaries u in $\mathrm {C}^*_u (X)$ . Let us show that, if the unitary $u\in \mathrm {C}^*_u (X)$ is properly chosen, then $\sigma _{uhu^*}$ is not coarse. For that, notice that, since h is unbounded, we can pick a vector $\xi \in \ell _2(X)$ with $\|\xi \|=1$ such that $\xi \not \in \operatorname {\mathrm {dom}} (h)$ . Moreover, without loss of generality, $\xi $ can be picked so that $\|\delta _{x_0}-\xi \|$ is arbitrarily small. Let $p=p_{x_0}$ and let $q $ be the rank $1$ projection onto the span of $ \xi $ . We can then assume that $ \|p-q\|$ is smaller than $1$ , so there is a unitary u in $\mathrm {C}^*_u (X)$ such that $uqu^*=p$ . Since $u\xi $ equals $\lambda \delta _{x_0}$ for some $\lambda $ in the unitary circle, replacing u by $\bar \lambda u$ , we can assume that $ u\xi =\delta _{x_0}$ . Hence, since $\xi $ is not in the domain of h, it follows that $\delta _{x_0}$ is not in the domain of $h'$ .

We can now conclude the proof. If $\sigma _{h'}$ were to be coarse, the operator $h'$ would be a coarse operator (Proposition 1.3.7). However, coarse operators have domains which are invariant under partial translations and, in particular, contain $c_{00}(X)$ . Since $\delta _{x_0}$ is not in $\operatorname {\mathrm {dom}} (h')$ , it follows that $\sigma _{h'}$ cannot be a coarse flow.

Proof of Theorem 1.4.5

Observing that each $\sigma _{h,t}$ clearly lies in the connected component of the identity (Definition 6.1.1), and also that $\sigma _{h,t}=\mathrm {Ad}(e^{ith})$ , it follows from Theorem 6.1.2 that $e^{ith}\in \mathrm {C}^*_u (X)$ .

Proof of Proposition 6.1.8

Suppose towards a contradiction that $f\colon X\to X$ is not close to the identity. Then there is a sequence $(x_n)_n$ in X such that

$$\begin{align*}\lim _{n\in \mathbb{N} }d(x_n,f(x_n))=\infty. \end{align*}$$

Let $A=\bigcup _{n\in \mathbb {N} }\{x_n\}$ . By hypothesis, there is a unitary $u\in \mathrm {C}^*_u (X)$ such that $\sigma (p_A)=up_Au^*$ . As automorphisms of uniform Roe algebras are compact preserving and strongly continuous (see [Reference Špakula and WillettŠW13, Lemma 3.1]), going to a subsequence if necessary, we can assume that

$$\begin{align*}\|e_{x_nx_n}u^*\sigma (e_{x_kx_k})u\|\leq 2^{-k-1} \end{align*}$$

for all distinct $k,n\in \mathbb {N} $ . Therefore,

(6.1.12) $$ \begin{align} \left \|e_{x_nx_n}u^*\sigma (e_{x_nx_n})u\right \|&\geq \left \|e_{x_nx_n}u^*\sigma (p_A)u\right \|- \sum _{k\neq n}\left \|e_{x_nx_n}u^*\sigma (e_{x_kx_k})u\right \| \\ &\geq \left \|e_{x_nx_n}p_A\right \|-\frac {1}{2}\notag \\ &\geq \frac {1}{2}\notag \end{align} $$

for all $n\in \mathbb {N} $ . Since $e_{x_nx_n}$ , $u^*\sigma (e_{x_nx_n})$ and $e_{f(x_n)f(x_n)}$ have rank $1$ , we have that

$$ \begin{align*}\left \|e_{x_nx_n}u^*\sigma (e_{x_nx_n})e_{f(x_n)f(x_n)}\right \|&=\left \|e_{x_nx_n}u^*\sigma (e_{x_nx_n})\right \|\left \|u^*\sigma (e_{x_nx_n})e_{f(x_n)f(x_n)}\right \|\\ &=\left \|e_{x_nx_n}u^*\sigma (e_{x_nx_n})u\right \|\left \|\sigma (e_{x_nx_n})e_{f(x_n)f(x_n)}\right \| \end{align*} $$

for all $n\in \mathbb {N} $ — this is a simple computation and it can be found, for instance, in [Reference Braga, Farah and VignatiBFV21, Lemma 6.5]. Therefore, it follows from (6.1.12) and our choice of f that

(6.1.13) $$ \begin{align} \left \|e_{x_nx_n}u^*\sigma (e_{x_nx_n})e_{f(x_n)f(x_n)}\right \| \geq \frac {\delta }{2} \end{align} $$

for all $n\in \mathbb {N} $ .

As $\sigma $ is an automorphism, we know from [Reference Braga and FarahBF21, Theorem 4.4] that there is $r>0$ such that each $\sigma (e_{x_nx_n})$ is $\delta /3$ -r-approximable. Since $u\in \mathrm {C}^*_u (X)$ , replacing r by a larger number if necessary, we can suppose that each $u^*\sigma (e_{x_nx_n})$ is $\delta /3$ -r-approximable. However, since $\lim _{n\in \mathbb {N} }d(x_n,f(x_n))=\infty $ , this contradicts (6.1.13). So, f must be close to the identity of X.

Proof of Theorem 6.1.7

Fix $t\in \mathbb {R} $ . By [Reference Baudier, Braga, Farah, Khukhro, Vignati and WillettBBF⁺22, Corollary 3.3], there is a map $f_t\colon X\to X$ such that

$$\begin{align*}\inf _{x\in X}\left \|\sigma _t(e_{xx})\delta _{f_t(x)}\right \|>0. \end{align*}$$

As $\sigma $ is in the connected component of the identity, there is a continuous path of projections in $\mathrm {C}^*_u(X)$ starting in $p_A$ and ending in $\sigma (p_A)$ . Therefore, it follows from [Reference MurphyMur90, Lemma 7.5.4] that $\sigma $ satisfies the hypothesis of Proposition 6.1.8. The result is now an immediate consequence of Proposition 6.1.8.

Proof of Theorem 6.1.2

By Theorem 6.1.7, there is a coarse equivalence $f\colon X\to X$ satisfying (6.1.4) which is close to the identity on X. Since the identity $X\to X$ is a bijection, applying Lemma 6.1.6 to $g=\mathrm {id}_X$ , we obtain that the unitary u implementing $\sigma $ must belong to $\mathrm {C}^*_u (X)$ as desired.

Proof of Theorem 7.1.3

By [Reference Špakula and ZhangŠZ20, Theorem 3.3], an operator $a\in \mathcal {B} (\ell _2(X))$ is in $\mathrm {C}^*_{\textit {ql}} (X)$ if and only if $[a,f]$ is compact for all Higson functions $f\colon X\to \mathbb {C} $ . Fix such f and let us show that $[e^{ih},f]$ is compact. This is equivalent to showing that $e^{ih}fe^{-ih}-f$ is compact, so, if $\pi \colon \mathrm {C}^*_u (X)\to \mathrm {C}^*_u (X)/\mathcal {K} (\ell _2(X))$ is the canonical quotient map, we must show that $\pi (e^{ih}fe^{-ih})=\pi (f)$ . For that, we prove that the map

(7.1.6) $$ \begin{align} t\in \mathbb{R} \mapsto \pi (e^{ith}fe^{-ith})\in \mathrm{C}^*_u (X)/\mathcal{K} (\ell _2(X)) \end{align} $$

is constant. Since this map equals $\pi (f)$ for $t=0$ , the result will follow.

Proof. As $h-h'$ is bounded on $c_{00}(X)$ and h satisfies (7.1.4), this implies that

$$\begin{align*}\sup \{\langle h'\delta _y,\delta _x\rangle \ :\ x,y\in X \text { with }x\neq y\}<\infty. \end{align*}$$

As $h'$ has finite propagation and

(7.1.8) $$ \begin{align} \langle [h',f]\delta _y,\delta _x\rangle =(f_y-f_x)\langle h'\delta _y,\delta _x\rangle \end{align} $$

for all $x,y\in X$ , the X-by-X matrix representation of $[h',f]$ has finite propagation and its coordinates are uniformly bounded. Since X is u.l.f., Proposition 2.3.1 implies that the X-by-X matrix $\big [\langle [h',f]\delta _y,\delta _x\rangle \big ]_{x,y\in X}$ canonically induces a bounded operator on $\ell _2(X)$ . Being a finite propagation operator, $h'$ is also assumed to be admissible. Based on the fact that f is a diagonal operator, one can easily prove that $c_{00}(X)$ is a core for $[h',f]$ . Therefore, since $[h',f]$ and the bounded operator induced by its matrix in the canonical basis of $\ell _2(X)$ clearly coincide on $c_{00}(X)$ , it follows that $[h',f]$ is itself bounded.

In order to show that $[h',f]$ is compact, it is enough to notice that

$$\begin{align*}\lim _{x,y\to \infty }\langle [h',f]\delta _y,\delta _x\rangle =0 \end{align*}$$

(here we use again that $[h',f]$ has finite propagation). But this follows immediately from (7.1.8) since f is Higson and $\operatorname {\mathrm {prop}} (h')<\infty $ .

Proof. By hypothesis, for every $\varepsilon>0$ , there is an operator $h'$ with finite propagation such that $h-h'$ is bounded and $\|h-h'\|<e$ . Since the equality

(7.1.10) $$ \begin{align} [h,f]=[h-h',f]+[h',f] \end{align} $$

holds on $c_{00}(X)$ , we have that

$$\begin{align*}\big \|[h,f]\big \|_0\leq \big \|[h-h',f]\big \|_0+\big \|[h',f]\big \|_0 \leq 2\varepsilon \|f\| + \big \|[h',f]\big \|_0 <\infty , \end{align*}$$

where by $\|\cdot \|_0$ we mean the norm of the restriction of the relevant operator to $c_{00}(X)$ . Again, since f is diagonal, $c_{00}(X)$ is a core for $[h,f]$ , so the above inequality implies that $[h,f]$ is bounded on its whole domain, as required. In addition,

$$\begin{align*}\big \|[h,f]-[h',f]\big \|_0 = \big \|[h-h',f] \big \|_0\leq 2\varepsilon \|f\|, \end{align*}$$

which means that $[h,f]$ may be approximated arbitrarily closely by compact operators, therefore itself being compact.

By Claim 7.1.9, $[h,f]$ is bounded. Therefore, [Reference Bratteli and RobinsonBR87, Proposition 3.2.55] implies that the map

(7.1.11) $$ \begin{align} t\in \mathbb{R} \mapsto e^{ith}fe^{-ith}\in \mathcal{B} (\ell _2(X)) \end{align} $$

is differentiable at zero and its derivative equals $i[h,f]$ . Moreover, since $\sigma _h$ is a group action, this implies that (7.1.11) is differentiable at all $t\in \mathbb {R} $ with derivative $\sigma _{h,t}(i[h,f])$ . As $[h,f]$ is compact (Claim 7.1.9) and as the canonical quotient map $\pi \colon \mathrm {C}^*_u(X)\to \mathrm {C}^*_u(X)/\mathcal {K}(\ell _2(X))$ is linear, we conclude that (7.1.6) is differentiable at all $t\in \mathbb {R} $ with derivative $0$ , hence a constant function. Consequently

$$\begin{align*}\pi (e^{ih}fe^{-ih}) = \pi (f), \end{align*}$$

whence $[e^{ih}, f]$ is compact, proving the required criterion for $e^{ih}$ to belong to $\mathrm {C}^*_{\textit {ql}} (X)$ .

For the last statement, notice that, by Proposition 4.1.13, any coarse operator satisfies (7.1.4).

Proof of Theorem 7.1.12

Let $\sigma _h$ be the pre-flow defined by h, which is actually a coarse flow thanks to Proposition 1.3.7. By Proposition 4.1.4, there is $s>0$ such that

$$ \begin{align*}\|[h,p_A] \|\leq s \end{align*} $$

for all $A\subseteq X$ . Hence, letting $\pi $ be the inclusion of $G=\{-1,1\}^X$ into $\ell _\infty (X)$ (see (7.1.13)), it follows that the family $([h,\pi (\varepsilon )])_{\varepsilon \in G}$ is uniformly bounded. Moreover, $[h,\pi (\varepsilon )]$ lies in $\mathrm {C}^*_u (X)$ , for every $\varepsilon $ in G. To see this, recall that each $[h,\pi (\varepsilon )]$ is given by the derivative of $\sigma _h$ :

$$\begin{align*}i[h,\pi (\varepsilon )]=\lim _{t\to 0}\frac {\sigma _{h,t}(\pi (\varepsilon ))-\pi (\varepsilon )}{t} \end{align*}$$

(see Corollary 5.1.4). So,

(7.1.18) $$ \begin{align} ([h,\pi (\varepsilon )])_{\varepsilon \in G}\in \ell _\infty (G,\mathrm{C}^*_u (X)). \end{align} $$

Notice that this family is also equi-approximable. For that, observe that

$$\begin{align*}\mathrm{SOT}\text {-} \sum _{x\in A}[h,p_{\{x\}}]=[h,p_A]\in \mathrm{C}^*_u (X) \end{align*}$$

for all $A\subseteq X$ . Lemma 6.1.11 then implies that $([h,\pi (\varepsilon )])_{\varepsilon \in G}$ is equi-approximable.

Fix an arbitrary $\theta>0$ , and let us show that there is an operator $h'$ on $\ell _2(X)$ , with finite propagation, and such that $h-h'$ is a bounded operator with norm at most $\theta $ . For that, using the equi-approximability of the family $([h,\pi (\varepsilon )])_{\varepsilon \in G}$ , pick $r>0$ , such that $[h,\pi (\varepsilon )]$ is $\theta $ -r-approximable for all $\varepsilon \in G$ . Since each $\pi (\varepsilon )$ has propagation zero, $\pi (\varepsilon )^*[h,\pi (\varepsilon )]$ is also $\theta $ -r-approximable for all $\varepsilon \in G$ . For each $\varepsilon \in G$ , pick $b_{\varepsilon }\in \mathrm {C}^*_u (X)$ with propagation at most r and such that

(7.1.19) $$ \begin{align} c_{\varepsilon }=\pi (\varepsilon )^*[h,\pi (\varepsilon )]-b_{\varepsilon } \end{align} $$

has norm at most $\theta $ . By (7.1.18), $([h,\pi (\varepsilon )])_{\varepsilon \in G}$ is bounded, so $(b_{\varepsilon })_{\varepsilon \in G}$ is bounded as well. Then, letting $\Psi $ be as in (7.1.16), we can define

$$\begin{align*}w=\Psi ((\pi (\varepsilon )^*[h,\pi (\varepsilon )])_{\varepsilon \in G}),\ b=\Psi ((b_{\varepsilon })_{\varepsilon \in G}),\ \text { and }\ c=\Psi ((c_{\varepsilon })_{\varepsilon \in G}). \end{align*}$$

Since each $b_{\varepsilon }$ has propagation at most r, Lemma 7.1.17 gives that the same holds for b. As each $c_{\varepsilon }$ has norm at most $\theta $ , so has c. Also, since we have (7.1.19) for all $\varepsilon \in \{-1,1\}^X$ , it follows that

(7.1.20) $$ \begin{align} c=w-b. \end{align} $$

Proof. First notice that, as w is bounded and h is a coarse operator, $w+h$ is also a coarse operator. Let now $x,y\in X$ be distinct and let us compute $\langle w\delta _y,\delta _x\rangle $ . Letting $e_{y,x}$ be the partial isometry of rank $1$ such that $e_{y,x}\delta _x=\delta _y$ , we have

$$ \begin{align*} \langle w\delta _y,\delta _x\rangle =\mathrm {Tr}(we_{y,x})=w(e_{y,x}), \end{align*} $$

where the last equality is given by (7.1.15). Therefore, by the formula of $\Psi $ , $\varphi $ , and using the duality relation in (7.1.15) again, we have

$$ \begin{align*} \langle w\delta _y,\delta _x\rangle &=\Psi ((\pi (\varepsilon )^*[h,\pi (\varepsilon )])_{\varepsilon \in G})(e_{y,x}) \\ &=\int _{\{-1,1\}^X} \mathrm {Tr}(\pi (\varepsilon )^*[h,\pi (\varepsilon )]e_{y,x}) d\varphi (\varepsilon ). \end{align*} $$

Notice that, even though h may not be bounded, the fact that $c_{00}(X)\subseteq \operatorname {\mathrm {dom}} (h)$ implies that $he_{y,x}$ is well defined and bounded. Therefore, we can write

$$\begin{align*}\pi (\varepsilon )^*[h,\pi (\varepsilon )]e_{y,x}=\pi (\varepsilon )^*h\pi (\varepsilon )e_{y,x}-he_{y,x} \end{align*}$$

and it follows that

$$ \begin{align*} \langle w\delta _y,\delta _x\rangle &=\int _{\{-1,1\}^X} \mathrm {Tr}(\pi (\varepsilon )^*h\pi (\varepsilon )e_{y,x})-\mathrm {Tr}(he_{y,x} )d\varphi (\varepsilon )\\ &=\int _{\{-1,1\}^X} \langle \pi (\varepsilon )^*h\pi (\varepsilon )\delta _y,\delta _x\rangle d\varphi (\varepsilon )-\langle h\delta _y,\delta _x\rangle. \end{align*} $$

Since x and y are distinct and $\langle E(h)\delta _y,\delta _x\rangle =0$ , Lemma 7.1.14 implies that the first term in the right-hand side above is zero. This shows that $\langle w\delta _y,\delta _x\rangle =-\langle h\delta _y,\delta _x\rangle $ and we conclude that

$$\begin{align*}\langle (w+h)\delta _y,\delta _x\rangle =0. \end{align*}$$

Since $x,y\in X$ were arbitrary distinct elements of X, this shows that $w+h$ has propagation zero as desired.

We can finally define $h'$ by letting

$$\begin{align*}h'=w+h-b. \end{align*}$$

First notice that, as b is bounded, and $w+h$ is a coarse operator, $h'$ is also a coarse operator. Moreover, since $w+h$ and b actually have finite propagation, so does $h'$ . By (7.1.20), we have that $h-h'=-c$ . So, $h-h'$ is a bounded operator with norm at most $\theta $ . As $\theta>0$ was arbitrary, this shows that $h\in \mathrm {C}^*_{u,unb}(X)$ .

For the second statement, notice that the first part of the theorem implies that h satisfies the hypothesis of Corollary 4.1.18. The result is then a consequence of this proposition.

Proof of Theorem 1.3.10

(2) $\Rightarrow $ (1): By [Reference Špakula and ZhangŠZ20, Theorem 3.3], since X has property A, we have $\mathrm {C}^*_u (X)=\mathrm {C}^*_{\textit {ql}} (X)$ . Hence, Theorem 7.1.3 gives that $\sigma _{h,t}(a)\in \mathrm {C}^*_u (X)$ for all $a\in \mathrm {C}^*_u (X)$ and, by Proposition 1.3.7, $\sigma _h$ is indeed a coarse flow.

(1) $\Rightarrow $ (2): By Proposition 1.3.7, h is a coarse operator and, by Theorem 7.1.12, $h\in \mathrm C^*_{u,unb}(X)$ .

(1) $\Rightarrow $ (3): By Theorem 7.1.12, h is in $\mathrm C^*_{u,unb}(X)$ and $h-E(h)$ is bounded. As h is in $\mathrm C^*_{u,unb}(X)$ , for each $\varepsilon>0$ , there is an operator $h'$ with finite propagation such that $h-h'$ has norm at most $\varepsilon $ . As h is a coarse operator (Proposition 1.3.7), the supremum of the off-diagonal coordinates of h is finite, hence the same is valid for $h'$ . In particular, as $h'-E(h')$ has finite propagation, it follows that $h'-E(h')$ is bounded (Proposition 2.3.1). Since