inductive limit of nuclear c*algebras is nuclear

Question

Let $(A_n,\varphi_n)_{n\in\mathbb{N}}$ be an inductive sequence of nuclear $C^*$-algebras $A_n$, i.e. $A_n\otimes_{min} C\cong A_n\otimes_{max} C$ for all $C^*$-algebras $C$.

How to prove that the inductive limit $\varinjlim (A_n,\varphi_n)=: A $ is nuclear?

I saw this statement in lecture notes, but there was only a hint and no proof. The hint is to prove it by case analysis:

1. Case: $\varphi_n$ is injective for all $n\in\mathbb{N}$.

2. Case: the general case. Use that for $0\to J\to B\to B/J\to 0$ (a short exact sequence of $C^*$-algebras) the $C^*$-algebra $B$ is nuclear iff $J$ and $B/J$ are nuclear.

However, I don't know how to prove it in detail. My ideas are:

For case 1, I guess that we can assume $A_n\subseteq A_{n+1}$ for all $n\in \mathbb{N}$, since the connecting maps $\varphi_n:A_n\to A_{n+1}$ are injective. By definition/construction of the inductive limit, $\hat{A}:=\bigcup\limits_{n\in\mathbb{N}}A_n$ is a dense subset of $A$. But I don't know how to continue.

[Edit Another hint for case 1 which I have seen in a book recently (but there was no proof, too) is: Use Arveson's extension theorem. I know this theorem, but I have no idea how to use it. ]

As regards case 2, consider $\hat{\varphi_n}:A_n/ker(\varphi_n)\to A_{n+1}$, $a\mapsto \hat{\varphi}([a]):=\varphi_n(a)$. Since $A_n$ is nuclear, $ker(\varphi_n) $ and $A_n/ker(\varphi_n)$ are nuclear ( the short exact sequence is $0\to ker(\varphi_n)\to A_n\to A_n/ker(\varphi_n)\to 0$). Furthermore, $\hat{\varphi_n}$ is injective for all $n\in\mathbb{N}$ and I guess it is $\varinjlim (A_n/ker(\varphi_n),\hat{\varphi_n})= A $. Then we can apply 1 and $A$ is nuclear.

What is a reference for a proof or can you help me to fill the gaps?

Regards

Answer 1

Since you mention Arveson's extension theorem, here is how one can use it to prove what you want. Nuclearity has an alternate definition

A $C^{\ast}$-algebra $A$ is nuclear iff there exists contractive completely positive (ccp) maps $\varphi_n : A\to M_{k(n)}(\mathbb{C})$ and $\psi_n : M_{k(n)}(\mathbb{C}) \to A$ such that $$ \lim_{n\to \infty}\|\psi_n(\varphi_n(a)) - a\| = 0 \quad\forall a\in A $$

It is then not too hard to prove (at least in the separable case) that

$A$ is nuclear iff, for each finite set $F \subset A$ and $\epsilon > 0, \exists n\in \mathbb{N}$ and ccp maps $\varphi : A\to M_n(\mathbb{C})$ and $\psi : M_n(\mathbb{C}) \to A$ such that $$ \|\psi(\varphi(a)) - a\| < \epsilon \quad\forall a\in F $$

This, together with Arveson's extension theorem can be weakened to a local property

$A$ is nuclear if, for each finite set $F \subset A$ and $\epsilon > 0, \exists$ a nuclear subalgebra $B\subset A$ such that, for each $a\in F, \exists b\in B$ with $$ \|a-b\| < \epsilon $$

Proof: Suppose $A$ satisfies the above condition, $F\subset A$ finite and $\epsilon >0$, then choose a nuclear subalgebra $B$ such that, for every $a\in F, \exists b\in B$ such that $$ \|a-b\| < \epsilon/3 \qquad (\dagger) $$ Let $G \subset B$ denote a finite set of such $b$. Then since $B$ is nuclear, but the above result, $\exists n\in \mathbb{N}$ and ccp maps $\varphi : B\to M_n(\mathbb{C})$ and $\psi : M_n(\mathbb{C}) \to B$ such that $$ \|\psi(\varphi(b)) - b\| < \epsilon/3 \quad\forall b\in G $$ Now, by Arveson's extension theorem, we extend $\varphi$ to a ccp map $\Phi : A\to M_n(\mathbb{C})$. Then, for any $a\in F$, choose $b\in B$ so that $(\dagger)$ holds, then \begin{equation*} \begin{split} \|\psi(\Phi(a)) - a\| &\leq \|\psi(\Phi(a)) - \psi(\Phi(b))\| + \|\psi(\Phi(b)) - b\| + \|b-a\| \\ &\leq \|a-b\| + \epsilon/3 + \epsilon/3 \\ &< \epsilon \end{split} \end{equation*} where the second inequality holds because both $\psi$ and $\Phi$ are contractive.

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It is then quite obvious that the inductive limit of nuclear $C^{\ast}$-algebras with injective connecting maps satisfies this last property.

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All this and more is covered in Brown and Ozawa's excellent book $C^{\ast}$-algebras and Finite dimensional approximations, which has pretty much everything one needs to know about nuclearity.