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<n, 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 nth 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

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