nuclear $C^*$ algebra

Question

If $(A_i)$ is a sequence of nuclear $C^*$ algebras,Is $\oplus_{c_0}A_i$ ($c_0$ direct sum)and $\prod A_i$($\ell ^\infty $ direct sum) also nuclear?

Answer 1

Fix $a\in \bigoplus_nA_n$ and $\varepsilon>0$. For each $n$, there exist ucp maps $\varphi_n:A_n\to M_{k(n)}(\mathbb C)$ and $\psi_\varepsilon:M_{k(n)}(\mathbb C)\to A_n$ such that $\|\psi_n\circ\varphi_n(a_n)-a_n\|<\varepsilon$. There is also $m$ such that $\|a_n\|<\varepsilon$ for all $n\geq m$. Write $a_0$ for the truncation of $a$ to its first $m$ entries; then $\|a-a_0\|<\varepsilon$. Then the maps $$ \varphi:\bigoplus_nA_n\to \bigoplus_{n=1}^mM_{k(n)}(\mathbb C),\ \ \ \psi:\bigoplus_{n=1}^mM_{k(n)}(\mathbb C)\to \bigoplus_nA_n $$ given by $$\varphi(b)=\bigoplus_{n=1}^m\varphi_n(b_n),\ \ \ \ \ \psi(\bigoplus_{n=1}^m c_n)=\bigoplus_{n=1}^m \psi_n(c_n)$$ satisfy \begin{align} \|\psi\circ\varphi(a)-a\| &\leq\|\psi\circ\varphi(a)-\psi\circ\varphi(a_0)\|+\|\psi\circ\varphi(a_0)-a_0\|+\|a_0-a\|\\ \ \\ &\leq 2\|a_0-a\|+\|\psi\circ\varphi(a_0)-a_0\|<3\varepsilon. \end{align} By making this work over the finite sets $F\subset \bigoplus_nA_n$ we obtain net $\{\varphi_F\}$ and $\{\psi_F\}$ such that $\|\psi_F\circ\varphi_F(a)-a\|\to0$ for all $a\in \bigoplus_nA_n$.

The diret product, on the other hand is not nuclear; it's not even exact. For instance $M=\prod_n M_n(\mathbb C)$ is not nuclear. It is well-known that nuclear algebras are exact, and that exactness passes to subalgebras. The full C$^*$-algebra of $\mathbb F_2$ is known to be non-exact, and to be residually finite; this means that there exists a faithful representation $\pi:C^*(\mathbb F_2)\to M=\prod_nM_n(\mathbb C)$. So $M$ cannot be exact, and in particular it is not nuclear.