Taken from C*-Algebras and Operator Theory by Gerald J. Murphy.

Although the principal aim of this section is to construct direct limits of C*-algebras, we begin with direct limits of groups.

If $\{G_n\}_{n=1}^\infty$ is a sequence of groups, and if for each $n$ we have a homomorphism $\varphi_n:G_n\rightarrow G_{n+1}$, then we call $\{G_n\}_{n\geq1}$ a direct sequence of groups. Given such a sequence and positive integers $n\leq m$, we set $\varphi_{nn}=I_{G_n}$ and we define $\varphi_{nm}:G_n\rightarrow G_m$ inductively on $m$ by setting

$\varphi_{n,m+1}=\varphi_m\varphi_{nm}$.

If $n\leq m\leq k$, we have $\varphi_{nk}=\varphi_{mk}\varphi_{nm}$.

If $G'$ is a group and we have homomorphisms $\theta^n:G^n\rightarrow G'$ such that the diagram

commutes for each $n$, that is $\theta^n=\theta^{n+1}\varphi_n$, then $\theta^n=\theta^m\varphi_{nm}$ for all $m\geq n$.

The product $\prod_{k=1}^\infty G_k$ is a group with the pointwise-defined operation, and if we let $G'$ be the set of all elements $(x_k)_{k\geq 1}$ in $\prod_{k\geq 1}G_k$ such that there exists $N$ for which $x_{k+1}=\varphi_k x_k$ for all $k\geq N$, then $G'$ is a subgroup of $\prod_{k\geq 1}G_k$ (this follows readily from the fact that the $\varphi_n$ are homomorphisms). Let $e_k$ be the identity of $G_k$. The set $F$ of all $(x_k)_{k\geq 1}\in \prod_{k\geq 1}G_k$ such that there exists $N$ for which $x_k=e_k$ for all $k\geq N$ is a normal subgroup of $G'$ (again from the homomorphism property), and we denote the quotient group $G'/F$ by $G$ (the cosets are sets of elements that have equal ‘tails’.). We call $G$ the direct limit of the sequence $\{G_n,\varphi_n\}_{n\geq 1}$, and where no ambiguity can result we sometimes write $\varinjlim G_n$ for $G$.

If $x\in G_n$, define $\hat{\varphi}^n(x)$ to be the sequence $(x_k)_{k\geq 1}$ where $x_k=e_k$ for $k, and $x_{n+k}=\varphi_{n,n_k}(x)$ for all $k\geq 0$ (tailed by the $\varphi_n$). Then $\hat{\varphi}^n(x)\in G'$, and the map

$\varphi^n:G_n\rightarrow G$$x\mapsto \hat{\varphi}^n(x)F$,

is a homomorphism, called the natural homomorphism from $G_n$ to $G$. It is straightforward to check that the diagram

commutes for each $n$, and that $G$ is the union of the increasing sequence $\{\varphi^n(G^n)\}_{n\geq 1}$.

## Theorem 6.1.1

Let $G$ be the direct limit of the direct sequence of groups $\{G_n,\varphi_n\}_{n\geq 1}$ and let $\varphi^n:G^n\rightarrow G$ be the natural map for each $n$.

1. If $x\in G_n$ and $y\in G_m$ and $\varphi^n(x)=\rho^m(y)$, then there exists $k\geq n,\,m$ such that $\varphi_{nk}(x)=\varphi_{mk}(y)$ (if the tails agree they’ll have to agree in ‘finite time’).
2. If $G'$ is a group, and for each $n$ there is a homomorphism $\theta^n:G_n\rightarrow G'$ such that the diagram
commutes, then there is a unique homomorphism $\theta:G\rightarrow G'$ such that for each $n$ the diagram
commutes.

Proof : Condition 1. follows directly from the definitions.

Assume we have $G'$ and $\theta^n$ as in condition 2. If $x\in G_n$ and $y\in G_m$ and $\varphi^n(x)=\varphi^m(y)$, then by condition 1. there exists $k\geq n,\,m$ such that $\varphi_{nk}(x)=\varphi_{mk}(y)$ . Hence,

$\theta^n(x)=\theta^k\varphi_{nk}(x)=\theta^k\varphi_{mk}(y)=\theta^m(y).$

Thus we can well-define a map $\theta:G\rightarrow G'$ by setting $\theta(\varphi^n(x))=\theta^n(x)$. It is easily checked that $\theta$ is a homomorphism, and by definition $\theta\varphi^n=\theta^n$ for all $n$. Uniqueness of $\theta$ is clear $\bullet$

### Remark

From condition 1. of the preceding theorem, if $\varphi^n(x)=e_{G'}$, then there exists $k\geq n$ such that $\varphi_{nk}(x)=e_{G_{k+1}}$, a result we shall be using frequently.

Let $A$ be a *-algebra. A C*-seminorm on $A$ is a seminorm $\rho$ on $A$ such that for all $a,\,b\in A$ we have

$\rho(ab)\leq\rho(a)\rho(b)$$\rho(a^*)=\rho(a)$ and $\rho(a^*a)=\rho(a)^2$.

If in addition $\rho$ is in fact a norm, we call $\rho$ a C*-norm.

If $\varphi:A\rightarrow B$ is a *-homomorphism from a *-algebra into a C*-algebra, then the function

$\rho:A\rightarrow \mathbb{R}^+$$a\mapsto\|\varphi(a)\|$,

is a C*-seminorm on $A$, and if $\varphi$ is injective, $\rho$ is a C*-norm.

If $\rho$ is any C*-seminorm on a *-algebra $A$, the set $N=\rho^{-1}\{0\}$ is a self-adjoint ideal of $A$ ($\checkmark$)m and we get a C*-norm on the quotient *-algebra $A/N$ by setting $\|a+N\|=\rho(a)$. If $B$ denotes the Banach space completion of $A/N$ with this norm, it is easily checked that the multiplication and involution operations extend uniquely to operations of the same type on $B$ so as to make $B$ a C*-algebra. We call $B$ the enveloping C*-algebra of the pair $(A,\rho)$, and the map

$i:A\rightarrow B$$a\mapsto a+N$,

the canonical map from $A$ to $B$. Of course, $i(A)$ is a dense *-subalgebra of $B$.

If $\rho$ is a C*-norm, we refer to $B$ more simply as the C*-completion of $A$. In this case $A$ is a dense *-subalgebra of $B$.

Let $\{A_n\}_{n\geq 1}$ be a sequence of C*-algebras and suppose that for each $n$ we have a *-homomorphism $\varphi_n:A_n\rightarrow A_{n+1}$. Then we call $(A_n,\varphi_n)_{n\geq 1}$ a direct sequence of C*-algebras. The product $\prod_{k\geq 1}A_k$ is a *-algebra with the pointwise-defined operations, and if $A'$ denotes the set of all elements $a=(a_k)_{k\geq 1}$ of $\prod_{k\geq 1}A_k$ such that there is an integer $N$ for which $a_{k+1}=\varphi_{k}(a_k)$ for all $k\geq N$, then $A'$ is a *-subalgebra of $\prod_{k\geq 1}A_k$. Note that $\|a_{k+1}\|\leq \|a_k\|$ if $k\geq N$ (since the $\varphi_n$ are norm-decreasing), so the sequence $\{\|a_k\|\}_{k\geq 1}$ is eventually decreasing (and of course bounded below). It therefore converges, and we set $\rho(a)=\lim_{k}\|a_k\|$. It is straightforward to verify that

$\rho:A'\rightarrow\mathbb{R}^+$$a\mapsto \rho(a)$,

is a C*-seminorm on $A'$. We denote the enveloping C*-algebra of $(A',\rho)$ by $A$, and call it the direct limit of the sequence $\{A_n,\varphi_n\}_{n\geq 1}$. If no ambiguity results, we sometimes write $\varinjlim A_n$ for $A$.

Similar to the group case, if $a\in A_n$, we define $\hat{\varphi}^n(a)$ in $A'$ to be the sequence $\{a_k\}_{k\geq 1}$ such that $a_1,\dots,a_{n-1}$ are zero and $a_{n+k}=\varphi_{n,n+k}(a)$ for all $k\geq 0$. If $i:A'\rightarrow A$ is the canonical map, then the map

$\displaystyle \varphi^n:A_n\rightarrow A$$\displaystyle a\mapsto i(\hat{\varphi}^n(a))$,

is a *-homomorphism, called the natural map from $A_n$ to $A$. A routine argument show that for all $n$ the diagram

commutes, and that the union of the increasing sequence of C*-subalgebras $\{\varphi^n(A_n)\}_{n\geq 1}$ is a dense *-subalgebra of $A$.

$\hat{\varphi}^n(a)\in A'$ is the element which is ‘evolves’ from $a$ in the $n$th place. That this diagram commutes is a very simple consequence of this.

Also,

$\displaystyle \left\|\varphi^n(a)\right\|=\lim_{k\rightarrow\infty}\left\|\varphi_{n,n+k}(a)\right\|$

if $a\in A_n$.

## Theorem 6.1.2

Let $A$ be the direct limit of the direct sequence of C*-algebras $\{A_n,\varphi_n\}_{n\geq 1}$, and suppose that $\varphi^n:A_n\rightarrow A$ is the natural map for each $n$.

1. If $a\in A_n$$b\in A_m$$\varepsilon>0$ and $\varphi^n(a)=\varphi^m(b)$, then there exists $k\geq n\,m$ such that $\|\varphi_{nk}(a)-\varphi_{mk}(b)\|<\varepsilon$.
2. If $B$ is a C*-algebra and there is a *-homomorphism $\psi^n:A_n\rightarrow B$ for each $n$ such that the diagram

commutes, then there is a unique *-homomorphism $\psi:A\rightarrow B$ such that for each $n$ the diagram

commutes.

Proof : If $\varphi^n(a)=\varphi^m(b)$ this implies that $\hat{\varphi}^n(a)-\hat{\varphi}^m(b)$ converges pointwise to $0$.

Suppose that $B$ and $\psi^n$ are as in condition 2. Let $a\in A_n$ and $b\in A_m$, and suppose that $\varphi^n(a)=\varphi^m(b)$. If $\varepsilon>0$, then by condition 1. there exists $k\geq n,\,m$ such that $\|\varphi_{nk}-\varphi_{mk}\|<\varepsilon$. Consequently,

$\|\varphi^n(a)-\varphi^m(b)\|=\|\psi^k(\varphi_{nk}(a)-\varphi_{mk}(b))\|\leq \|\varphi_{nk}(a)-\varphi_{mk}(b)\|<\varepsilon$.

Letting $\varepsilon\rightarrow0$, we therefore have $\psi^n(a)=\psi^m(b)$. This shows that we can well-define a map from $\psi$ to

$\displaystyle \bigcup_{n=1}^\infty \varphi^n(A_n)$ to $B$

by setting $\psi(\varphi^n(a))=\psi^n(a)$. If $k$ is any integer, then

$\|\psi^n(a)\|=\|\psi^{n+k}\varphi_{n,n+k}(a)\|\leq \|\varphi_{n,n+k}(a)\|$,

and therefore

$\displaystyle \|\psi(\varphi^n(a))\|=\|\psi^n(a)\|\leq\lim_{k\rightarrow\infty}\|\varphi_{n,n+k}(a)\|=\|\varphi^n(a)\|$.

Thus $\psi$ is norm-decreasing, and it it easily seen to be a *-homomorphism. Since $C$ is a dense *-subalgebra of $A$, we can extend $\psi$ to a *-homomorphism $\psi:A\rightarrow B$, and $\psi\varphi^n=\psi^n$ for all $n$. Uniqueness follows from the density of $C$ in $A$ $\bullet$

### Remark

Retaining the notation of the preceding theorem, if $a\in A_n$ and $\varphi^n(a)=0$ and $\varepsilon>0$, then by condition 1. there exists $k\geq n$ such that $\|\varphi_{nk}(a)\|<\varepsilon$.

### Remark

Let $A$ be a C*-algebra and let $\{A_n\}_{n\geq 1}$ be an increasing sequence of C*-subalgebras of $A$ whose union is dense in $A$. Let $\varphi_n:A_n\rightarrow A_{n+1}$ be the inclusion map. A straightforward application of Theorem 6.1.2 condition 2., shows that $A$ is *-isomorphic to the direct limit of the direct sequence $\{A_n,\varphi_n\}_{n\geq 1}$.

## Theorem 6.1.3

Let $S$ be a non-empty set of simple C*-subalgebras of a C*-algebra $A$. Suppose that $S$ is upwards-directed (that is if $B,\,C\in S$, then there exists a $D\in S$ such that $B\cup C\subset D$), and $\displaystyle \bigcup S$ is dense in $A$. Then $A$ is simple also.

Proof : To show that $A$ is simple it suffices to show that if $\pi:A\rightarrow B$ is a surjective *-homomorphism onto a non-zero  C*-algebra $B$, then it is injective ($\pi$ is linear and non-zero, so if it not injective, the kernel is non-zero so that $A$ is not simple.). If $C\in S$, then the restriction of $\pi$ to $C$ is either $C$, or it is injective, and therefore isometric. Since $\pi$ is not the zero map on $\displaystyle \bigcup S$, it follows easily from the upwards-directed property of $S$ that $\pi$ is not the zero map on any non-zero $C\in S$. Hence, $\pi$ is isometric on $\displaystyle S$, and therefore, by continuity, $\pi$ is isometric on $A$ $\bullet$

## Theorem 6.1.4

Suppose that $\{A_n,\varphi_n\}_{n\geq 1}$ is a direct sequence of simple C*-algebras. Then the direct limit $\varinjlim A_n$ is simple also.

Proof : Let $\varphi^n:A_n\rightarrow A$ be the natural map, where $A=\varinjlim A_n$. Then the set $S=\{\varphi^n(A_n):n\geq 1\}$ is an upwards-directed family of simple C*-subalgebras of $A$ whose union is dense in $A$, so by Theorem 6.1.3, $A$ is simple $\bullet$