The proof of "Every quasidiagonal C*-algebra is stably finite"

Question

Here is a quotation in a book "C*-algebras and finite-Dimensional Approximations" by Nate and Taka (P241).

Recall that an isometry $s$ is called proper if $1-ss^{*}\neq0$

Definition 7.1.14 A unital C*-algebra $A$ is stably finite if $M_{n}(\mathbb{C})\otimes A$ contains no proper isometries, for every $n\in \mathbb{N}$.

Proposition 7.1.15 Every QD (quasidiagonal) C*-algebra is stably finite.

Proof. The proof boils down to the following routine exercise: If $T_{n}\in M_{k(n)}(\mathbb{C})$ and $||T_{n}^{*}T_{n}-1_{k(n)}||\rightarrow 0$, then $||T_{n}T_{n}^{*}-1_{k(n)}||\rightarrow 0$ too.

Hoping for a contradiction, we assume $A$ is quasidiagonal and that it contains a proper isometry $s\in A$. (Since $M_{n}(A)$ is QD whenever $A$ is, we may assume $A$ contains the proper isometry.) Let $\phi_{n}: A\rightarrow M_{k(n)}(\mathbb{C})$ be asymptotically multiplicative, asymptotically isometric u.c.p. maps. Then $1=\phi_{n}(s^{*}s)\approx\phi_{n}(s^{*})\phi(s)$ while $\phi_{n}(s)\phi_{n}(s)^{*}$ can not get close to 1 since $1-ss^{*}\neq 0$. This contradicts the exercise above.

My question are :

1. If $A$ is quasidiagonal, how to verify $M_{n}(A)$ is also quasidiagonal?

2. In the proof, why does the author assume $A$ contains a proper isometry, not $M_{n}(A)$ contains a proper isometry? (It has relations with $M_{n}(A)$ is QD? )

3. In the proof, does $\phi_{n}(s^{*})=\phi_{n}(s)^{*}$ hold? I mean how to get that $\phi_{n}(s)\phi_{n}(s)^{*}$, not $\phi_{n}(s)\phi_{n}(s^{*})$, can not get close to 1?

Answer 1

1. Suppose that $A$ is quasi-diagonal, that is, $A$ embeds into $B=\prod M_k / \sum M_k$. It is enough to show that $M_n ( B ) $ embeds into $B$. This however is clear. We can identify elements $X = ( [ [X^k_{ij}] ]_{i,j=1,\ldots n})_{k=1}^\infty $ of $M_n(B)$ with sequences in $B$ which are zero apart from terms whose index is divisible by $n$. More specifically consider the following map $\Phi \colon M_n(B)\to B$, $\Phi(X) = [(Y_n)_{n=1}^\infty]$ which is a well-defined injective *-homomorphism (please check!):

$$Y_n = \left\{ \begin{array}{ll} 0, & m\not|\,n, \\ [ X^k_{ij} ]_{i,j=1,\ldots n} & n = km \text{ for some }k. \end{array}\right.$$

2. Recall that $A$ is stably finite if $M_n(A)$ does not contain any proper isometries for all $n$. This is a proof by contradiction.

3. $\phi_n$ is completely positive, so it commutes with taking adjoints.