## The bounded coarse structure on a metric space is the topological coarse structure associated to its Higson compactification.

This post is rehearsal about Proposition 2.47 in Lectures in Coarse Geometry by Roe. Any comment is appreciated as well as usage of language. PROPOSITION 2.47. Let $(X, d)$ be a proper metric space. Then the bounded coarse structure on $X$ is the topological coarse structure associated to its Higson compactification. Poof.$\newcommand{\d}{\operatorname{d}}$ We only need to verify that the bounded coarse structure on $X$ is coarser than the topological coarse structure associated to its Higson compactification. Suppose $E\subset X\times X$ is continuously controlled by $\operatorname{h} X$ but not boundedly controlled, then for each $n$ there is a pair $(x_n,y_n)\in E$ with $d(x_n,y_n)>2n$. Without loss of generality, suppose $x_n$ tends to infinity of $X$. Then $y_n$ tends to infinity of $Y$ since   $E$ is continuously controlled. Fix a point $O$ in $X$. Because $X$ is a proper metric space (that is, any  bounded set is relatively compact), we can choose a subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that $x_{n_1}=x_1$ and  $x_{n_{k+1}}$ is out of the closed ball $B(O, r_k)$ where $r_k=\max\{d(O, x_{n_k}), d(O,y_{n_k})\}+2k+2$. It is easy to see that  closed balls $B(x_{n_k}, k)$ are disjoint pairwise. Define $f(x)=\begin{cases}\frac{k-d(x,x_{n_k})}{k} & \mbox{if}~ d(x,x_{n_k})\sup_{(x,y)\in F}d(x,y)$ and  $(x,y)$ is an arbitrary point in $F$. - $d(x, x_{n_k})<k$ and $d(y, x_{n_k})<k$ for some $k$.  $|\d f(x,y)|=\frac{|d(x,x_{n_k)}-d(y,x_{n_k})|}{k}\leq \frac{d_0}{k}.$ - $d(x, x_{n_k})<k$ for some $k$ while $d(y, x_{n_k})\geq k$. $0\leq f(x)=\frac{k-d(x,x_{n_k})}{k}\leq \frac{d(y,x_{n_k})-d(x,x_{n_k})}{k}\leq \frac{d_0}{k}.$ If $d(y, x_{n_j})<j$ for some $j\neq k$, then $|\d f(x,y)|\leq \frac{2d_0}{\min \{j,k\}}.$ If $d(y, x_{n_j})\geq j$ for all $j$, then $|\d f(x,y)|\leq \frac{d_0}{k}.$ - $d(x, x_{n_k})\geq k$ and $d(y, x_{n_k})\geq k$ for all $k$.  $f(x)=f(y)=0$, so $\d f(x,y)=0$. Let $B$ be the closed ball $B(O, r_k)(\supset \cup_{i=1}^k B(x_{n_i},i))$, then $K:=(B\times \overline{F^{-1}[B]})\cup (\overline{F[B]}\times B)$ is a compact set of $X\times X$. If $(x,y)\in F\backslash K$, then $x\notin B$ and $y\notin B$, thus $|\d f(x,y)|\leq \frac{2d_0}{k+1}.$ Therefore, $f\in C_h(X)$. Hence $\d f(x_n,y_n)$ tends to zero as $(x_n,y_n)$ tends to infinity of $X\times X$. However, $\d f(x_{n_k}, y_{n_k})=-1$ by the definition of $f$, a contradiction.

## Solutions to Some Exercises in Murphy’s book (Chapter 3)

$\newcommand{\zhang}[1]{{\Large{\text{#1}}}} \def \A{\mathscr A} \def \F{\mathbb{F}} \def \N{\mathbb{N}} \def \R{\mathbb{R}} \def \C{\mathbb{C}}\def \H{\mathscr{H}}\def \d{\mathrm{d}} \def \Inv{\mathrm{Inv}} \def \ni{^{-1}} \def \jx{\lim_{n\to\infty}} \def \yb{\frac{1}{2}} \newcommand{\x}{\times} \newcommand{\wt}{\textbf{Question.}} \newcommand{\bh}[1]{\textbf{#1.}} \newcommand{\xl}[2]{#1_1, #1_2, \cdots, #1_{#2}} \newcommand{\xl}[2]{#1_1, #1_2, \cdots, #1_{#2}} \newcommand{\tm}[1]{\textbf{ #1.}} \renewcommand{\l}[1]{\lVert #1\rVert}\newcommand{\zm}{Proof.} \newcommand{\zb}{}$ $\bh{3.5}$An element a of $A_+$ is strictly positive if the hereditary $C^*$-subalgebra of $A$ generated by $a$ is $A$ itself, that is, if $(aAa)^- = A$. (a) Show that if $A$ is unital, then $a \in A_+$ is strictly positive if and only if $a$ is invertible. (b) If $H$ is a Hilbert space, show that a positive compac operator on $H$ is strictly positive in $K(H)$ if and only if it has dense range. (c) Show that if $a$ is strictly positive in $A$, then $\tau( a) > 0$ for all non-zero positive linear functionals $\tau$ on $A$. $\zm$ (a) If $a$ is invertible, then $aAa=A$, thus $a$ is strictly positive. Conversely, if $a$ is strictly positive, then $(aAa)^-=A$. Since the set $G$ of all invertible elements is open, $aAa\cap G\neq\emptyset$. Hence there exists some $b\in A$ such that $aba$ is invertible, thus $a$ is invertible. (b)If $u$ is strictly positive in $K(H)$, that is $(uK(H)u)^-=K(H)$, then $\forall x\in H,$ there exists a sequence $\{v_n\}$ in $B(H)$ such that $$x\otimes x=\lim_{n\to \infty} uv_nu.$$ Hence $$\l{x}^2x=x\otimes x(x)=\lim_{n\to\infty}u(v_nu(x))\in (uH)^-.$$ Therefore, $x\in (uH)^-$ and $H=(uH)^-$. Conversely, if $(uH)^-H$, then for all $x\in H,$ then there exists a sequence $\{h_n\}$ in $H$ such that $uh_n\to x$, hence $$u(h_n\otimes h_n )u= (uh_n)\otimes(uh_n)\to x\otimes x.$$ Therefore, every rank-one projection belongs to $(uK(H)u)^-$. Since all the rank-one projections is the total subset of $K(H)$, $K(H)=(uK(H)u)^-$. (c)Suppose $\tau$ is a positive functional such that $\tau(a)=0$, then $$\tau((a^{\frac{1}{2}})^*a^\yb)=0,$$ so $$\tau(a^* a)=\tau((a^\frac{1}{2})^3 a^\yb)=0.$$ Hence for all $b\in A$, $$\tau(aba)=0.$$ Therefore, $\tau=0$ for $A=(aAa)^-$. $\bh{3.3}$If $\varphi:A\to B$ is a positive linear map between $C^*$ algebra, show that $\varphi$ is bounded. $\zm$ For all positive linear functional $\tau$ on $B$, $\tau\circ\varphi$ is a positive linear functional on $A$ and thus bounded, i.e.,$$\lVert\tau\circ\varphi(x)\rVert\leq M_\tau ~~ (\forall x\in A \text{ with } \l{x}\leq 1)$$ for some positive constant $M_\tau$. Every bounded linear functional on $B$ is a linear combination of 4 positive linear functional on $B$ by Jordan Decomposition, thus for all $b^*\in B^*,$ $$\l{b^*(\varphi(x))}\leq M_{b^*} ~~( \forall x\in A \text{ with } \l{x}\leq 1)$$ for some positive constant $M_{b^*}$. Hence $\l{\varphi(x)}\leq M$ for some positive constant $M$ by the Principle of Uniform Boundedness, i.e. $\varphi$ is bounded.

## left maximal modular ideal in $M_n$

Each pure state on $M_n=B(\mathbb{C}^n)$ is of form $$w_x:=\langle \cdot ~ x, x\rangle,$$ thus each left maximal modular ideal is of form $$\{u|w_x(u^*u)=0\}=\{u|\|ux\|=0\}.$$ Particularly, corresponding to $e_1=(1,0)$ in $\mathbb{C}^2$, the left maximal modular ideal is is of form $$\left( \begin{array} & 0 & a\\ 0 & b \end{array}\right).$$